Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, the limit is
$$2\lim\limits_{x\to+\infty}\frac{x\sqrt{1-\frac{1}{x}}\times t}{x^n}$$
What to do next, if this is a good approach?
| Correction to your answer: $$2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x^2}}$$
Your approach is going to have trouble because $t$ is a fairly complicated mess.
Note:
$$\frac{x+\sqrt{x^2-1}}{x} = 1+\sqrt{1-\frac{1}{x^2}}\to 2\text{ as } x\to +\infty$$
and
$$\frac{x-\sqrt{x^2-1}}{x}= 1-\sqrt{1-\frac{1}{x^2}}\to 0\text{ as }x\to+\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| The final decimal digit of $a^2$ and the final decimal digit of $2b^2$ can't agree unless $a$ and $b$ are both multiples of five, leading to an infinite descent. (Check: the possible last digits of $a^2$ are 0,1,4,5,6,9 and the possible last digits of $2b^2$ are 0,2,8.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 3
} |
Prove that $|U|^2=2[\cosh \sqrt{2 m}- \cos \sqrt{2 m}]$ How can we prove that $|U|^2=2[\cosh \sqrt{2 m}- \cos \sqrt{2 m}]$
When $U=e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)}$
I expand
$e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)}=e^{\sqrt{\frac{m}{2}}}[\cos \sqrt{\frac{m}{2}}+i \sin\sqrt{\frac{m}{2}}] -e^{-\sqrt{\frac{m}{2}}}[\cos \sqrt{\frac{m}{2}}-i \sin\sqrt{\frac{m}{2}}]$
$=\cos \sqrt{\frac{m}{2}}[e^{\sqrt{\frac{m}{2}}}-e^{-\sqrt{\frac{m}{2}}}]+i \sin\sqrt{\frac{m}{2}}[e^{\sqrt{\frac{m}{2}}}+e^{-\sqrt{\frac{m}{2}}}]$
I don't know the next step from here on.
| $|U|^2=U\times U^*$, in which $U^*$ means replacement of $i$ with $-i$.
$|U|^2=U\times U^*=(e^{\sqrt{\frac{m}{2}}(1+i)}-e^{-\sqrt{\frac{m}{2}}(1+i)})\times (e^{\sqrt{\frac{m}{2}}(1-i)}-e^{-\sqrt{\frac{m}{2}}(1-i)})= e^{\sqrt{\frac{m}{2}}(2)}-e^{\sqrt{\frac{m}{2}}(+2i)}-e^{\sqrt{\frac{m}{2}}(-2i)}+e^{\sqrt{\frac{m}{2}}(-2)}=2[\cosh \sqrt{2 m}- \cos \sqrt{2 m}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Absolute Value Inequality
$$|\frac{x-1}{x+3}|\leq 2$$
I solve it as follow:
$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{3}=-\frac{5}{3}\leq x\leq -7$
I now see that I did not take into consideration $x\neq -3$ which can not be
The book on the other side says that in answer is $x\leq -7$ or $x\geq -\frac{5}{3}$
Am I wrong?
| Just break it into sections by sign:
$\left|\frac{x-1}{x+3}\right| =
\left\{
\begin{array}{ll}
\frac{x-1}{x+3} & \mbox{if } x \geq 1 \\
-\frac{x-1}{x+3} & \mbox{if } x \in (-3,1) \\
\frac{x-1}{x+3} & \mbox{if } x \lt -3
\end{array}
\right.$
Then solve the three inequalities based on the sign of $x+3$:
$$x\geq1:\frac{x-1}{x+3} \leq 2 \iff x \geq -7$$
$$x\in(-3,1):-\frac{x-1}{x+3} \leq 2 \iff x \geq -\frac{5}{3}$$
$$x<-3:\frac{x-1}{x+3} \leq 2 \iff x \leq -7$$
That gives us the ranges $[1,\infty)$, $[-\frac53,1)$, and $(\infty,-7)$. So the final solution is $x \in (\infty,-7) \cup [-\frac53,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Give me a hint to solve this integral $$\int_0^1 \frac{1}{(x^2+x+1)^2} dx$$
I don't have any clue as to where to even start. I am struggling on this one.
Please shed some light by giving me a Hint
| $$\dfrac{d\left[\dfrac{ax^2+bx+c}{x^2+x+1}\right]}{dx}$$
$$=\dfrac{(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
We need $1=(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)=(a-b)x^2+(a-2c)x+b-c$
$\implies a-b=0,a-2c=0,b-c=1\implies \cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Closed form for the sum $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$ I want a closed form for the sum $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$
where: $a\ne 1<p<n;\quad p\in\mathbb Z$
I know a related identity,
$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1-\frac{p}{2}\right)\left\lfloor\frac{n}{p}\right\rfloor-\frac{p}{2}\left\lfloor\frac{n}{p}\right\rfloor^2\quad\;,$$
but I can't apply it to this problem!
Result for this sum:
$$\quad\displaystyle S=\dfrac{a^{n+1}}{a-1}\left\lfloor\frac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}$$
Can you prove it?
| My Solution
Apply the Summation by parts, we have:
\begin{align*} S=\sum_{k=1}^n a^k\left\lfloor\dfrac{k}{p}\right\rfloor&=\dfrac{1}{a-1}\sum_{k=1}^n\left\lfloor\dfrac{k}{p}\right\rfloor\Delta(a^k)
\\ &= \left.\dfrac{1}{a-1}\cdot a^k\left\lfloor\dfrac{k}{p}\right\rfloor\right|_{k=1}^{n+1}-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right) \\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right)\end{align*}
Note that: $\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor=\begin{cases}1,\quad p\mid k+1\\0,\quad p\nmid k+1\end{cases}=\begin{cases}1,\quad k=mp-1\\0,\quad otherwise\end{cases}$
Therefore:
\begin{align*}S&=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{1\le mp-1\le n} a^{mp}\\ &=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \dfrac{a^{p(m+1)}-a^{pm}}{a^p-1}
\\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{(a-1)(a^p-1)}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \Delta[a^{pm}]
\\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\left.\dfrac{a^{pm}}{(a-1)(a^p-1)}\right|_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor+1}\\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?
| I would use complex exponentials. We have:
\begin{align}
\sin^{24}\frac{\pi}{24}+\cos^{24}\frac{\pi}{24}&=
\left(\frac{e^{i\frac{\pi}{24}}-e^{-i\frac{\pi}{24}}}{2i}\right)^{24}
+\left(\frac{e^{i\frac{\pi}{24}}+e^{-i\frac{\pi}{24}}}{2}\right)^{24}\\
&=\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} (-1)^k e^{i \frac{24-2k}{24}\pi}\right)
+\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} e^{i \frac{24-2k}{24}\pi}\right)\\
&= \frac{1}{2^{23}}\sum_{l=0}^{12} \binom{24}{2l} e^{i \frac{12 - 2l}{12}\pi}
\end{align}
after canceling every other term in the two sums. We can rewrite this last expression in terms
of trig functions again, as
$$
\frac{1}{2^{22}} \left(\sum_{m=0}^{5} \binom{24}{2m} \cos \left(\pi-\frac{m\pi}{6}\right) +
\frac{1}{2} \binom{24}{12}\right)
$$
(here we're folding the sum in half and taking advantage of the fact that
$\binom{n}{k}=\binom{n}{n-k}$, which is why the middle term is anomalous).
Now, as $\cos x=-\cos(\pi - x)$, this collapses to a reasonable number of terms:
$$
\frac{1}{2^{22}}\left[\frac{1}{2}\binom{24}{12} - 1
+ \left(\binom{24}{10} - \binom{24}{2}\right) \cos \frac{\pi}{6}
+ \left(\binom{24}{8} - \binom{24}{4}\right) \cos \frac{\pi}{3}
\right]
$$
which simplifies into
$$
\frac{1}{2^{22}} \left(\frac{3428999}{2} + 1960980\frac{\sqrt{3}}{2}\right)
=\frac{3428999 + 1960980 \sqrt{3}}{2^{23}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 2
} |
Explanation of Generating Functions Simplification Given the problem:
What is the coefficient of $x^{2005}$ in the generating function $G(x) = \frac{1}{(1-x)^2(1+x)^2}$?
The solution posted starts with:
Let $\frac{1}{(1-x)^2(1+x)^2} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{C}{1+x} + \frac{D}{(1+x)^2}$. Upon simplification, the right hand side becomes:
$$\frac{(C-A)x^3 + (B+D-A-C)x^2 + (A+2B-C-2D)x + (A+B+C+D)}{(1-x)^2(1+x)^2}$$
What step-by-step solution did he follow in order to come up with this kind of ordered simplification? I'm sure he used extended binomial theorem to do this but can anyone give me a proper explanation how he applied that to this problem?
| We have
\begin{align}
&\frac A{1-x} + \frac B{(1-x)^2} + \frac C{1+x} + \frac D{(1+x)^2}\\
&=\frac{A(1-x)(1+x)^2 + B(1+x)^2 + C(1-x)^2(1+x) + D(1-x)^2}{(1-x)^2(1+x)^2}\\
&=\frac{A(1+x-x^2-x^3) + B(1+2x+x^2) + C(x^3-x^2-x+1) + D(1-2x+x^2)}{(1-x)^2(1+x)^2}\\
&=\frac{(-A+C)x^3 + (-A+B-C+D)x^2 + (A+2B-C-2D)x + (A+B+C+D)}{(1-x)^2(1+x)^2}.
\end{align}
and hence
$$(-A+C)x^3 + (-A+B-C+D)x^2 + (A+2B-C-2D)x + (A+B+C+D) = 1$$
From this we get the system of linear equations
\begin{align}
-A+C &= 0\\
-A+B-C+D&=0\\
A+2B-C-2D&=0\\
A+B+C+D&=1
\end{align}
which yields
$$ A=B=C=D = \frac14.$$
Thus the generating function is
\begin{align}
&\frac14\sum_{n=0}^\infty x^n + \frac14\sum_{n=0}^\infty (n+1)x^n + \frac14\sum_{n=0}^\infty (-1)^n x^n + \sum_{n=0}^\infty(-1)^n (n+1)x^n\\
&= \sum_{n=0}^\infty \frac14 \left(1 + (-1)^n\right)(n+2)x^{2n}\\
&= \sum_{n=0}^\infty (n+1)x^{2n}
\end{align}
and clearly the coefficient of $x^{2005}$ is zero.
A much easier way to solve this problem, though, is to recognize that
$$(1-x)^2(1+x)^2 = ((1-x)(1+x))^2 = (1-x^2)^2. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inverse Laplace Transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. I am trying to figure out the inverse laplace transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. Looking at my table of Laplace Transforms in my textbook, it seems that either I must break up this fraction using partial fractions into linear terms or I need to use some trick that I'm missing. I know that the denominator cannot be broken up further into linear terms, since if I try to solve $s^2 - 6s + 13$ I get complex roots, so I'm unsure what to do. Any help is appreciated.
| Given:
$$Y(s)=\frac{6s-19}{s^2-6s+13}$$
APPROACH #1:
Completing the square of the denominator gives
$$Y(s)=\frac{6s-19}{(s-3)^2+4}=\frac{6s}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}$$
Algebraic manipulation of each term gives an equivalent equation in a more appropriate form (reasons behind this step will become apparent shortly)
$$Y(s)=6\frac{s-3+3}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}=6\frac{s-3}{(s-3)^2+4}-\frac{1}{(s-3)^2+4}$$
or
$$Y(s)=6\frac{s-3}{(s-3)^2+4}-\frac{1}{2}\frac{2}{(s-3)^2+2^2}$$
From the transform tables, we use the pairs:
*
*$e^{at}cos(bt)\leftrightarrow \frac{s-a}{(s-a)^2+b^2}$
*$e^{at}sin(bt)\leftrightarrow \frac{b}{(s-a)^2+b^2}$
Identifying: $a=3$ and $b=2$, we get
$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} = 6e^{3t}cos(2t)-\frac{1}{2}e^{3t}sin(2t)$$
APPROACH #2:
Invoking the idea of partial fraction decomposition, we write
$$\frac{6s-19}{s^2-6s+13}=\frac{A}{s-3-2i}+\frac{B}{s-3+2i}$$
or after multiplying by $s^2-6s+13$
$$(s-3+2i)A+(s-3-2i)B=6s-19$$
*
*If $s=3-2i$:
$$0+(3-2i-3-2i)B=6(3-2i)-19\Rightarrow B=3-\frac{1}{4}i$$
*If $s=3+2i$:
$$(3+2i-3+2i)A+0=6(3+2i)-19\Rightarrow A=3+\frac{1}{4}i$$
Therefore
$$Y(s)=\frac{3+\frac{1}{4}i}{s-3-2i}+\frac{3-\frac{1}{4}i}{s-3+2i}$$
From the transform tables, we use the pair:
*
*$\alpha e^{(a+jb)t}\leftrightarrow \frac{\alpha}{s-(a+jb)}$
to invert the terms.
Identifying: $a=3$ and $b=2$, we get
$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} = (3+\frac{1}{4}i)e^{\left (3+2i \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i \right )t}$$
To show that this is equivalent to the one found above, we use little complex algebra:
\begin{align}
y(t) &= (3+\frac{1}{4}i)e^{\left (3+2i \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i \right )t} \\
& = e^{3t}\left [ (3+\frac{1}{4}i)e^{2it}+(3-\frac{1}{4}i)e^{-2it} \right ] \\
& = e^{3t}\left [ 3(e^{2it}+e^{-2it})+\frac{1}{4}i(e^{2it}-e^{-2it})\right ]\\
& = e^{3t}\left [ 3(2cos(2t))+\frac{1}{4}i(2isin(2t))\right ] \\
& = e^{3t}\left [ 6cost(2t)-\frac{1}{2}sin(2t)\right ] \\
& = 6\;e^{3t}cost(2t)-\frac{1}{2}\;e^{3t}sin(2t)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Combination sum . I want to evaluate the following sum :
$$S(k,k')=\sum_{i} C_{i+k}^k C_{k'-i}^{k}$$ = $$S(k,k')=\sum_{i} \binom{i+k}{k} \binom{k'-i}{k}$$
I tried some steps but couldnt get further than :
$S(k,k')=2*(\sum_{i}^{(k'-k)/2}C_{i+k}^k C_{k'-i}^{k})$
| This yields to snake oil (see e.g. Wilf's "generatingfunctionology").
We want a special case of "Vandermonde on its head":
$$
\sum_k \binom{n - k}{r} \binom{m + k}{s}
$$
We will need the identity:
$$
\sum_k \binom{k}{n} z^k
= \frac{z^n}{(1 - z)^{n + 1}}
$$
and negative binomial coefficients:
$$
\binom{-n}{k}
= (-1)^k \binom{n + k - 1}{n - 1}
$$
Choose $n$ as free variable, so our sum is $s_n$, and consider the generating function:
$\begin{align}
S(z) &= \sum_{n \ge 0} s_n z^n \\
&= \sum_{n \ge 0} z^n \sum_k \binom{n - k}{r} \binom{m + k}{s} \\
&= \sum_k \binom{m + k}{s} \sum_{n \ge 0} \binom{n - k}{r} z^n \\
&= \sum_k \binom{m + k}{s} z^k \sum_{n \ge 0} \binom{n - k}{r} z^{n - k} \\
&= \sum_k \binom{m + k}{s} z^k \sum_j \binom{j}{r} z^j \\
&= \sum_k \binom{m + k}{s} \frac{z^k \cdot z^r}{(1 - z)^{r + 1}} \\
&= \frac{z^{r - m}}{(1 - z)^{r + 1}} \sum_k \binom{m + k}{s} z^{m + k} \\
&= \frac{z^{r - m}}{(1 - z)^{r + 1}} \cdot \frac{z^s}{(1 - z)^{s + 1}} \\
&= \frac{z^{r + s - m}}{(1 - z)^{r + s + 2}}
\end{align}$
Now we want:
$\begin{align}
s_n
&= [z^n] S(n) \\
&= [z^n] \frac{z^{r + s - m}}{(1 - z)^{r + s + 2}} \\
&= [z^{m + n - r - s}] \frac{1}{(1 - z)^{r + s + 2}} \\
&= (-1)^{m + n - r - s} \binom{-r - s - 2}{n + m - r - s} \\
&= \binom{r + s + 2 + m + n - r - s - 1}{r + s + 2 - 1} \\
&= \binom{m + n + 1}{r + s + 1}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant by Gauss Elimination I want to find the determinant by Gauss Elimination Method for the following matrix:
$$\begin{bmatrix}
1 & 2 & 0 & 1\\
0 & -1 &1 & -4\\
1 & 2 & 0 & -2\\
2 & 0& 5 &8
\end{bmatrix}$$
My attempt is
Row3-row1 and Row4-2Row1
$$\begin{bmatrix}
1 & 2 & 0 & 1\\
0 & -1 &1 & -4\\
0 & 0 & 0 & -3\\
0 & -4& 5 &6
\end{bmatrix}$$
the next step is
Row4-4Row2
$$\begin{bmatrix}
1 & 2 & 0 & 1\\
0 & -1 &1 & -4\\
0 & 0 & 0 & -3\\
0 & 0& 1 &22
\end{bmatrix}$$
What is the next step??
| The operations you performed don't change the determinant; you can also swap two rows, which however multiplies the determinant by $-1$.
Swap rows 3 and 4, you matrix becomes triangular and its determinant is
$$
1\cdot(-1)\cdot 1\cdot(-3)=3
$$
so the given matrix has determinant $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c$, $a^2+b^2+c^2$, $a^3+b^3+c^3$ are real, then so are a,b,c
Let $a,b,c$ be complex numbers with distinct magnitudes such that $a+b+c$, $a^2+b^2+c^2$, $a^3+b^3+c^3$ are real.
Prove that $a,b,c$ are real numbers as well.
I tried to go for contradiction: WLOG assume $a$ is not real. Since $a+b+c\in \mathbb R$, at least one of the numbers $b$ or $c$ is not real. WLOG $b$ is not real. I couldn't get any contradiction from that.
I also attempted something using elementary symmetric polynomials: let $s=a+b+c$, $q=ab+ac+bc$ and $p=abc$
Since $a^2+b^2+c^2=s^2-2q$, $\;\;q$ is real as well.
Since $a^3+b^3+c^3=s^3-3qs+3p$, $\;\;p$ is real.
What then?
Source: an oral exam at a French engineering school.
| Hint: As you've shown, $(x-a)(x-b)(x-c)$ has real coefficients. What can be said about the complex roots of a polynomial with real coefficients?
A more geometric proof would be to first show that if any of $a,b,c$ are real, we can replace that value with any real value and it is still true, so you can assume that $|a|<|b|<|c|=1$ and $c$ is not real, by scaling and replacing any real values with smaller values.
But we can prove inductively (using your realization that $p,q$ are real) that $a^n+b^n+c^n$ is always real. Then pick $N$ so that $|a|^n+|b|^n<\epsilon$ for $n>N$. Then the imaginary part of $c^n$ would have to be less than $\epsilon$ for all $n>N$. Then show this is impossible for a particular $\epsilon$ unless $c$ is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solving recurrence relation without initial condition Any idea on how I can approach this recurrence relation? It is very different to other questions I have encountered where there is only one term of $T(n)$ on the RHS, and the initial condition isn't given as well.
$$T(n)=\frac2n\big(T(0)+T(1)+\ldots+T(n-1)\big)+5n$$
for $n\ge 1$.
| Use generating functions. Define $t(z) = \sum_{n \ge 0} T(n) z^n$, shift the recurrence and multiply out to get rid of fractions; multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
(n + 1) T(n + 1)
&= \sum_{0 \le k \le n} T(k) + 5 (n + 1)^2 \\
\sum_{n \ge 0} (n + 1) T(n + 1) z^n
&= \sum_{n \ge 0} z^n \sum_{0 \le k \le n} T(k)
+ 5 \sum_{n \ge 0} (n + 1)^2 z^n \\
\frac{1}{z} \sum_{n \ge 0} n T(n) z^n
&= \sum_{n \ge 0} z^n \sum_{0 \le k \le n} T(k)
+ 5 \frac{1 + z}{(1 - z)^3} \\
\frac{1}{z} \cdot z t'(z)
&= \frac{t(z)}{1 - z} + 5 \frac{1 + z}{(1 - z)^3}
\end{align*}$
No need for any initial values here. Solve for $t(z)$:
$\begin{align*}
t(z)
&= \frac{10}{(1 - z)^2} + 5 \frac{1}{1 - z} \ln \frac{1}{1 - z} + \frac{c}{1 - z}
\end{align*}$
Using the generalized binomial theorem and the generating function of the harmonic numbers:
$\begin{align*}
(1 - z)^{-m}
&= \sum_{n \ge 0} (-1)^n \binom{-m}{n} z^n \\
&= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} z^n \\
H_n
&= \sum_{1 \le k \le n} \frac{1}{n} \\
\sum_{n \ge 0} H_n z^n
&= \frac{1}{1 - z} \ln \frac{1}{1 - z}
\end{align*}$
we see that:
$\begin{align*}
T(n)
&= [z^n] t(n) \\
&= 10 \binom{n + 2 - 1}{2 - 1} + 5 H_n + c \\
&= 10 n + 5 H_n + c'
\end{align*}$
Here $c' = T(0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this indefinite integral with $\arctan$? $$\int \frac{(x^2-1)\;\text{d}x}{(x^4+3x^2+1) \tan^{-1}{\left(\frac{x^2+1}{x}\right)}}$$
We should divide numerator and denominator by $ x^2 $ and put $z=x+\frac{1}{x}$ but I'm still not getting the answer. Please help!
| Consider the integral:
\begin{align}
I(a) = \int \frac{(x^2 - 1) \, dx}{(x^4 + a x^2+ 1) \, \tan^{-1}\left(\frac{x^2+1}{x}\right) }.
\end{align}
The integral can be seen in the form
\begin{align}
I(a) = \int \frac{\left(\frac{x^2-1}{x^2}\right) \, dx}{\left( a -1 + \left( x + \frac{1}{x} \right)^2 \right) \, \tan^{-1}\left(x + \frac{1}{x} \right)}.
\end{align}
Make the substitution $u = x + \frac{1}{x}$ to obtain
\begin{align}
I(a) &= \int \frac{du}{(a-1 + u^2) \, \tan^{-1}(u)}.
\end{align}
Making the substitution $t = \tan^{-1}(u)$ leads to
\begin{align}
I(a) = \int \frac{\sec^{2}(t) \, dt}{(a-1 + \tan^{2}(t)) \, t} = \int \frac{dt}{t \, (1 + (a-2) \cos^{2}(t))}.
\end{align}
This result shows that if $a \neq 2$ then the integral has no closed form result. If $a=2$ then
\begin{align}
I(2) &= \ln(t) = \ln(\tan^{-1}(u)) = \ln\left(\tan^{-1}\left(x + \frac{1}{x}\right)\right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Continuously changing Dimensions of a Rectangle The dimension of a rectangle are continousl changing. The width increases at a rate of 3 in/s while the length decreases at the rate of 2 in/s. At one instant the rectangle is a 20-in square. How fast is its area changing 3 seconds later?
The answer is -16.
$$\begin{align}A &= LW,\, 20 = LW \\[1ex]
\frac{dA}{dt} &= L\frac{dW}{dt} + W\frac{dL}{dt}\\[1ex]
\frac{dA}{dt} &= \frac{20}W\cdot 3 + W\cdot(-2)\\[2ex]
t=3\implies W &= (3\text{ in/s})*3\text s = 9\text { in}\\[1ex]
\frac{dA}{dt} &= \frac{20}9\cdot3 + 9\cdot(-2) = \frac{34}3\end{align}$$
Is my assumption at t=3 wrong? any hint?
| "The rectangle is a 20-in square" means that it is a $20\times 20$ square; that is, $L(0) = W(0) = 20$. Three seconds later, $L(3) = 20-3\cdot 2 = 14$ and $W(3) = 20 + 3\cdot 3 = 29$, so that
$$
\frac{dA}{dt}\bigg\lvert_{t=3}
= L(3)\frac{dW}{dt}\bigg\lvert_{t=3} + W(3)\frac{dL}{dt}\bigg\lvert_{t=3}
= 14\cdot 3 + 29(-2) = -16.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How does one construct the Galois field extension $GF((2^2)^3)$? Looking at past exam question, one asks us to construct a Galois field extension $GF((2^2)^3)$ whenever the primitive irreducible polynomial $p(X) = X^3 + \alpha X^2 + \alpha X + \alpha \in GF(2^2)[X]$ is given. How do you answer this?
| Ok. It is easy to check that $p(X)$ is irreducible. Let $\beta$ be a zero of $p(X)$. Then
$$
\begin{aligned}
\beta^3&=\alpha\beta^2+\alpha\beta+\alpha,\\
\beta^4&=\beta^2+\beta+\alpha^2,\\
\beta^5&=\alpha^2\beta^2+\beta+\alpha,\\
\beta^6&=\alpha^2\beta+1.
\end{aligned}
$$
Now we can observe that $\beta^6+\beta^5+\beta^3=\beta^2+1$, so the minimal polynomial of $\beta$ over $GF(2)$ is $m(X)=X^6+X^5+X^3+X^2+1$.
At this point it is easy to cheat, fire up Mathematica, and check that $m(X)$ is not a factor of $x^{21}+1$. The only sextic factor of $X^9+1$ is $X^6+X^3+1$, so we now know that $\beta$ is of order 63, i.e. a primitive element.
Let's start squaring (using what we have calculated already) for a change to verify this
$$
\begin{aligned}
\beta^8&=(\beta^4)^2=\beta^4+\beta^2+\alpha^4&=\beta+1,\\
\beta^{16}&=(\beta^8)^2&=\beta^2+1,\\
\beta^{32}&=(\beta^{16})^2=\beta^4+1&=\beta^2+\beta+\alpha,\\
\beta^{64}&=(\beta^{32})^2=\beta^4+\beta^2+\alpha^2&=\beta.
\end{aligned}
$$
Well, well! There's a fair chance that no errors crept in, given that $\beta^{64}$ checks out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Resolve this system: Im tried to resolve this problem:
$$\max\quad f\left( x,y \right) =xy\quad \text{s.a}\quad \begin{cases} x^2 +y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
Well, i form the lagrangian and the respective gradient, so i had this system to resolve:
$$\begin{cases} y+2\lambda_1 x+\lambda_2=0 \\ x+2\lambda_1 y+ \lambda_2=0 \\ x^2+y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$
And i can't find all the solutions, i need help-
| $z=-(x+y), x^2+y^2+(x+y)^2=1 \implies x^2+y^2+xy=\dfrac{1}{2},x^2+y^2\ge 2xy \implies 3xy \le \dfrac{1}{2} \iff xy\le \dfrac{1}{6}$,
when $x=y \cap x^2=\dfrac{1}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Nesbitt's Inequality for 4 Variables I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.
Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds:
$$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$
I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).
My effort: I understood how the author defines the variables $M, N$ and $S$.
$$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$
$$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$
$$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$
$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.
Then the author, without any expansion/explanation, says
$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$
Which is also true, since the AM-GM inequality says
$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$
The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.
The next part is the crux of my question.
The author claims,
$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$
This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??
As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.
| Since we have $(x-y)^2\ge 0$, we have, for $x\gt 0,y\gt 0$,
$$\begin{align}(x-y)^2\ge 0&\Rightarrow x^2+y^2+2xy\ge 4xy\\&\Rightarrow y(x+y)+x(x+y)\ge 4xy\\&\Rightarrow \frac{1}{x}+\frac 1y\ge\frac{4}{x+y}\end{align}$$
Now set $x=b+c,y=a+d$ and $x=c+d,y=a+b$ to get
$$\frac{1}{b+c}+\frac{1}{a+d}\ge\frac{4}{b+c+a+d}$$and$$\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{c+d+a+b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Limit of a complicated function. Find $$\lim\limits_{x \to 2^{-}} \frac{e^{((x+2)\log 4){\frac{[x+1]}{4}}}-16}{ 4^x -16}$$ where $[x]$ denotes the greatest integer function less than or equal to x.
ATTEMPT:
I tried the following substitution $x=2-h$ and as $x \to 2$, $h \to 0.$ which gave me something like this
$\lim\limits_{h\to 0}$$ e^{((4-h)ln4){\frac{[3-h]}{4}}}-16\over \frac{16}{4^h} -16.$The denominator seems solvable as we get $\lim\limits_{h\to 0}$ $-16\frac{(4^h-1)h}{4^h*h}$ which seems a standard limit.But the numerator stills remain the same.
How can i simplify the numerator?
| If we assume that there is a typo in the question (as mentioned in my comment) then (The mentioned typo has been fixed by OP)
We can write the given function as $$f(x) = \dfrac{\exp\left\{\{(x + 2)\log 4\}\cdot\dfrac{[x + 1]}{4}\right\} - 16}{4^{x} - 16}$$ When $x \to 2^{-}$ we can write $[x + 1] = 2$ and then the expression for $f(x)$ can be simplified to $$f(x) = \dfrac{\exp\left\{\left(1 + \dfrac{x}{2}\right)\log 4\right\} - \exp(2\log 4)}{4^{x} - 16}$$ Putting $x = 2 - h$ where $h \to 0^{+}$ we can see that the desired limit $L$ can be calculated as follows
\begin{align}
L &= \lim_{h \to 0^{+}}\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4\right\} - \exp(2\log 4)}{4^{2 - h} - 16}\notag\\
&= \exp(2\log 4)\lim_{h \to 0^{+}}\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{\dfrac{16}{4^{h}} - 16}\notag\\
&= \frac{\exp(2\log 4)}{16}\lim_{h \to 0^{+}}4^{h}\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{1 - 4^{h}}\notag\\
&= 1\cdot\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{h}\cdot\frac{h}{1 - 4^{h}}\notag\\
&= -\frac{1}{\log 4}\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{h}\notag\\
&= -\frac{1}{\log 4}\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4}\cdot\dfrac{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4}{h}\notag\\
&= -\frac{1}{\log 4}\lim_{t \to 0^{-}}\frac{e^{t} - 1}{t}\cdot\left(-\frac{\log 4}{2}\right)\notag\\
&= \frac{1}{2}
\end{align}
In the above derivation we have used substitution $$t = \left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4$$ and $t \to 0^{-}$ when $h \to 0^{+}$. Also the following standard limits are used $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1, \lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Height of the Trapezoid
Problem: The area of a trapezoid is equal to 2 and the sum of his diagonals is equal to 4. Find the trapezoid height.
[QUESTION]: I find a result that implies that the height of the Triangle is not uniquely defined, any help discussing this result or with other solutions is appreciated
Attempt: So using the notation of the figure I have:
$$a(\square ABCD)=2,\: \: \: AC+BD=d_1+d_2=4, \: \: h=?$$
So I make the following construction mirroring the trapezoid twice, once vertically and once horizontally:
Here I have put both trapezoid together sharing the $\overline{BC}$ segment.
Clearly $a(\triangle DB\hat{A})=a(\square ABCD)=\Big(\frac{b_1+b_2}{2}\Big)h=2$
Using Heron's Formula:
$$a(\triangle DB\hat{A})=\sqrt{s(s-(b_2+b_1))(s-d_1)(s-d_2)}$$
whre $s=\frac{b_1+b_2+d_1+d_2}{2}$
Using $B=(b_2+b_1)$ for short :
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(B+d_1+d_2)(-(B)+d_1+d_2)(B-d_1+d_2)(B+d_1-d_2)}$$
Using the beginning relations :
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(B+4)(-(B)+4)(B-(d_1-d_2))(B+(d_1-d_2))}$$
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(-B^2+16)(B^2-(d_1-d_2)^2))}$$
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{-B^4-B^2(d_1-d_2)^2+16B^2-16(d_1-d_2)^2)}$$
Now replacing $a(\triangle DB\hat{A})=2$
$$64=-B^4+B^2(16-(d_1-d_2)^2)-16(d_1-d_2)^2$$
$$64=-B^4+B^2(16-(4-2d_2)^2)-16(4-2d_2)^2$$
I can solve this Polynomial in Wolfram (where I changed $B\rightarrow y$ and $d_2 \rightarrow x$), obtaining many solutions
$$B = \pm\Big(\sqrt{2} \sqrt{-d_2^2+\sqrt{(d_2-2)^2 (d_2^2-4d_2-20)}+4 d_2}\Big)$$
$$B = \pm\Big(\sqrt{2} \sqrt{-d_2^2-\sqrt{(d_2-2)^2 (d_2^2-4d_2-20)}+4 d_2}\Big)$$
This yields that I can't find a unique value of $B$ and hence of $h$, which sort of make sense because I can find many triangles $\triangle DB\hat{A}$ that match the problem conditions of Area and sum of diagonals. This is kind of as if one have the Base and the height of a triangle but one cannot specify the side lengths.
I don't Know if there is another thing I can use, any help is appreciated.
| There is a way to find $h$, though quite tedious.
By Heron's formula,
$$ 2=\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)},$$
where
$ s=\frac{\hat{A}B+BD+\hat{A}D}{2}=\frac{4+\frac{4}{h}}{2}=2+\frac{2}{h}.$
\begin{eqnarray}
2&=&\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)}
\\&=& \sqrt{\left(2+\frac{2}{h}\right)\left(2+\frac{2}{h}-\hat{A}B\right)\left(2+\frac{2}{h}-BD\right)\left(2+\frac{2}{h}-\frac{4}{h}\right)}
\\&=&\sqrt{\left(2+\frac{2}{h}\right)\left(\left(2+\frac{2}{h}\right)^2-\left(2+\frac{2}{h}\right)(\hat{A}B+BD)+\hat{A}BBD\right)\left(2-\frac{2}{h}\right)}
\\&=& \sqrt{\left(4-\frac{4}{h^2}\right)\left(\left(2+\frac{2}{h}\right)^2-4\left(2+\frac{2}{h}\right)+\hat{A}BBD\right)}
\end{eqnarray}
Let $k=\hat{A}BBD$.
Back to Heron:
\begin{align}
4 &= &\left(4-\frac{4}{h^2}\right)\left(\left(2+\frac{2}{h}\right)^2-4\left(2+\frac{2}{h}\right)+k\right)
\\ 4h^4&=&\left(4h^2-4\right)\left(\left(2h+2\right)^2-4\left(2h^2+2h\right)+kh^2\right)
\\ &=&\left(4h^2-4\right)\left(-4h^2+4+kh^2\right)
\\ &=&4(k-4)h^4-4(k-4)h^2+16h^2-16 \end{align}
$$(4k-20)h^4+(32-4k)h^2-16=0 \tag{*}\label{*}$$
The discriminant $b^2-4ac\geq 0$ as $h^2$ is real.
\begin{align}
(32-4k)^2-4(4k-20)(-16)&\geq& 0
\\ 1024-256k+16k^2+256k-1280 &\geq& 0
\\ 16k^2-256&\geq& 0 \end{align}
Thus, $k \geq 4 $ or $ k \leq -4 $. Since $k$ is positive, $k \geq 4$.
$\hat{A}BBD=k$ and $\hat{A}B + BD =4$. Hence we now introduce quadratic equation
$$x^2-4x+k=0. $$
Again, $b^2-4ac \geq 0$ since $\hat{A}B$ and $BD$ are real. Then
\begin{align}
16-4k \geq 0
\\ k \leq 4
\end{align}
Then $4 \leq k \leq 4$, which means that $k=4$.
Therefore, substituting $4$ for $k$ in $\eqref{*}$ brings us $h=\sqrt{2}$.
Appendix
If $h$ can be found, then the statement that $h$ cannot be determined is not true. The problem in the argument is that some clues are neglected such as the use of discriminant. Actually, $d_2$ can be determined if we find the discriminant of the (quadratic) Polynomial with respect to $B^2$. We will know that $d_2=2$ by some geometric interpretations, and hence that $B=2\sqrt2$. $B$ can be determined, so can $h$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the fourier series of the function Find the fourier series of the function
$g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)}$ for $x \not= k\pi$, and $g(k\pi) = \lim_{x\to k\pi} g(x)$, $(k \in \mathbb{Z})$
| The problem is stated to be
\begin{align}
g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)} \mbox{ for } x \not= k\pi, \mbox{ and } g(k\pi) = \lim_{x\to k\pi} g(x), \hspace{5mm} k \in \mathbb{Z}
\end{align}
Consider the following.
\begin{align}
g(x,t) &= \sum_{n=1}^{\infty} \frac{\sin(nx) \, t^{n}}{\sin(x)} \\
&= \sum_{n=1}^{\infty} \frac{t^{n} \, ( \cos(2x) \, \sin((n-2)x) + \sin(2x) \, \cos((n-2)x)}{\sin(x)} \\
&= - \cos(2x) \, t + \cos(2x) \, \sum_{n=2}^{\infty} \frac{t^{n} \, \sin((n-2)x)}{\sin(x)} + 2 \, \cos(x) \, \sum_{n=1}^{\infty} t^{n} \, \cos((n-2)x) \\
&= - \cos(2x) \, t + \cos(2x) \, t^{2} \, g(x,t) + 2 \cos^{2}(x) t + 2 \cos(x) \, t \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n} \\
&= t + \cos(2x) \, t^{2} \, g(x,t) + 2 \, t \, \cos^{2}(x) \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n}
\end{align}
This leads to
\begin{align}
g(x,t) = \frac{t}{1 - \cos(2x) \, t^{2}} \left[ 1 + 2 \cos^{2}(x) \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n} \right].
\end{align}
In general
\begin{align}
g(x,t) &= \frac{t}{1 - \cos(2x) \, t^{2}} \left[ 1 + \frac{2 \cos^{2}(x) \, (1 - \cos(x) \, t)}{1 - 2 \cos(x) \, t + t^{2}} \right] \\
&= \frac{t}{1 - \cos(2x) \, t^{2}} \cdot \frac{1 + 2 \cos^{2}(x) - 2 \cos x (1 + \cos^{2} x) \, t}{1 - 2 \cos x \, t + t^{2}}.
\end{align}
For the case $x \to k \pi$ then
\begin{align}
g(k \pi, t) = \frac{t}{1 - t^{2}} \cdot \frac{3 - 4 (-1)^{k} \, t}{1 - 2 (-1)^{k} \, t + t^{2}}
\end{align}
When $t = 1/6$ this becomes
\begin{align}
g\left(k \pi, \frac{1}{6}\right) = \frac{72}{35} \, \frac{9 - 2(-1)^{k}}{37 - 12 (-1)^{k}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate an Integral Evaluate:
$$\int_{0}^{\infty}\dfrac{\sin^3(x-\frac{1}{x} )^5}{x^3} dx$$
I've been stumped by this Integral and cannot think of how to evaluate it.
I substituted $\dfrac{1}{x^2}=t \Rightarrow \dfrac{-dt}{2}= \dfrac{dx}{x^3}.$
$$$$However, I can't understand what to do next. $$$$Any help on solving this would be truly appreciated. Many thanks!
| as @achille hint,I have post it
since
$$I=\int_{0}^{+\infty}\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=(\int_{0}^{1}+\int_{1}^{\infty})\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=I_{1}+I_{2}$$
since
$$I_{2}=-\int_{0}^{1}x\sin^3{(\dfrac{1}{x}-x)^5}dx$$
so
$$I=\int_{0}^{1}\left(\dfrac{1}{x^3}-x\right)\sin^3{\left(x-\dfrac{1}{x}\right)^5}dx=\int_{0}^{\infty}=\int_{0}^{1}\left(\dfrac{1}{x}-x\right)\sin^3{\left(x-\dfrac{1}{x}\right)^5}d\left(x-\dfrac{1}{x}\right)$$
let $x-\dfrac{1}{x}=t$,then
$$I=-\int_{0}^{+\infty}u\sin^3{u^5}du=-\dfrac{1}{5}\int_{0}^{+\infty}\sin^3{u}u^{-\frac{3}{5}}du=-\dfrac{1}{20}\int_{0}^{+\infty}(3\sin{u}-\sin{3u})u^{-3/5}du$$
and use well known result
$$\int_{0}^{+\infty}\sin{x}\cdot x^pdx=\cos{(\pi p/2)}\Gamma{(1+p)},-2<p<0$$
can find it.becasue
$$-\dfrac{3}{20}\int_{0}^{+\infty}\sin{u}\cdot u^{-\frac{3}{5}}du=-\dfrac{3}{20}\cos{\dfrac{6\pi}{5}}\Gamma{(\dfrac{2}{5})}$$
we know
$$\cos{\dfrac{6\pi}{5}}=-\cos{\dfrac{\pi}{5}}=-\sqrt{1-\sin^2{\dfrac{\pi}{5}}}=-\dfrac{1}{2}\sqrt{\dfrac{5+\sqrt{5}}{2}}$$
and other is simaler
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how to decompose $\binom{x+y}{n}$ and $\binom{xy}{n}$ as a sum of products of $\binom{x}{0},\dots, \binom{x}{n}$ and $\binom{y}{0},\dots,\binom{y}{n}$. This can be done in principle by peeling off the binomials starting with the highest degree and working one's way down. Playing around with Sage, one soon guesses that
$\binom{x+y}{n} = \sum_{k=0}^n \binom{x}{k}\binom{y}{n-k}$
and in fact, I think this has a combinatorial proof which straightforwardly generalizes that of the identity $\binom{m}{n} = \binom{m}{n-1} + \binom{m-1}{n-1}$.
But $\binom{xy}{n}$ seems to be not so straightforward. The first few expansions are:
$\binom{xy}{2} = 2\binom{x}{2}\binom{y}{2} + x\binom{y}{2} + y \binom{x}{2}$
$\binom{xy}{3} = 6\binom{x}{3} \binom{y}{3} + $
$\qquad ~ ~ 6 \binom{x}{3}\binom{y}{2} + 6\binom{x}{2}\binom{y}{3} + $
$\qquad ~ ~ x \binom{y}{3} + 4 \binom{x}{2} \binom{y}{2} + y \binom{x}{3} $
$\binom{xy}{4} = 24\binom{x}{4}\binom{y}{4} + $
$\qquad ~ ~ 36\binom{x}{3}\binom{y}{4} + 36 \binom{x}{4}\binom{y}{3} + $
$ \qquad ~ ~ 14 \binom{x}{2}\binom{y}{4} + 45\binom{x}{3}\binom{y}{3} + 14 \binom{x}{4}\binom{y}{2} + $
$\qquad ~ ~ 12 \binom{x}{2}\binom{y}{3} + 12 \binom{x}{3}\binom{y}{2} + $
$\qquad ~ ~ \binom{x}{2}\binom{y}{2}$
and it is not so easy to discern a pattern.
This must be well-known: what is a closed-form expression for the expansion of $\binom{xy}{n}$?
| The identity $\binom{x+y}{n} = \sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$ is well-known, it is called the Vandermonde identity. The answer for $\binom{xy}{n}$ can be explained using the notion of a $\lambda$-ring, where here we consider the binomial ring $\mathbb{Z}$ with $\lambda^n(x)=\binom{x}{n}$. The main theorem on symmetric polynomials enables us to write
$$\sum_{k=0}^{n} P_k(\sigma_1,\dotsc,\sigma_k,\tau_1,\dotsc,\tau_k) \,t^k = \prod_{i,j=1}^{n} (1+ t x_i y_j)$$
for some polynomials $P_k \in \mathbb{Z}[x_1,\dotsc,x_k,y_1,\dotsc,y_k]$, where $\sigma_1,\dotsc,\sigma_n$ are the elementary symmetric polynomials in $x_1,x_2,\dotsc,x_n$ and $\tau_1,\dotsc,\tau_n$ are the elementary symmetric polynomials in $y_1,\dotsc,y_n$. Then, one has (by definition) $$\lambda^n(xy)=P_n\bigl(\lambda^1(x),\dotsc,\lambda^n(x),\lambda^1(y),\dotsc,\lambda^n(y)\bigr).$$
For example:
$$\lambda^2(xy) = x^2 \lambda^2(y) + \lambda^2(x) y^2 - 2 \lambda^2(x) \lambda^2(y)$$
$$\lambda^3(xy)=x^3 \lambda^3(y) + \lambda^3(x) y^3 + x \lambda^2(x) y \lambda^2(y) - 3 x \lambda^2(x) \lambda^3(y) - 3 \lambda^3(x) y \lambda^2(y) + 3 \lambda^3(x) \lambda^3(y)$$
Of course, one has to prove that every binomial ring becomes a $\lambda$-ring. See for instance Darij Grinberg's notes, Theorem 7.2 (which is a corollary of Theorem 7.1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A tough integral:$\int_0^{+\infty}\left( \frac1{\log(x+1)-\log x}-x-\frac12\right)^2 dx$ I would like to prove the convergence of
$$I=\int_0^{+\infty}\left( \frac1{\log(x+1)-\log x}-x-\frac12\right)^2 dx$$
then obtain a closed form of $I$.
Convergence is ensured by the fact that $x \mapsto f(x)=\left( \frac1{\log(x+1)-\log x}-x-\frac12\right)^2$ is continuous on $(0,+\infty)$ with $f(x) \sim \dfrac14$ as $x \to 0^+$ and $f(x) \sim \dfrac1{144 x^2}$ as $x \to +\infty.$
I did not succeed in finding a closed form of $I$. My attempt was to consider a certain parameter integral then make some differentiation to get rid of the $\log$ terms...
| We have:
$$ I = \frac{1}{4}\int_{0}^{+\infty}\left(\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t}\right)^2\frac{dt}{\sinh^2(t/2)}$$
where:
$$\frac{1}{2}+\frac{1}{e^t-1}-\frac{1}{t} = \sum_{n\geq 1}\frac{2t}{t^2+4\pi^2 n^2}\tag{1}$$
as well as:
$$ \frac{1}{\sinh^2(t)}=\frac{1}{t^2}+2\sum_{n\geq 1}\frac{(t^2-n^2\pi^2)}{(t^2+n^2\pi^2)^2}\tag{2}$$
so our integral depends on the two series:
$$ \sum_{n\geq 1}\sum_{m\geq 1}\int_{0}^{+\infty}\frac{dt}{(t^2+4n^2\pi^2)(t^2+4m^2\pi^2)}=\frac{1}{16\pi^2}\sum_{n\geq 1}\sum_{m\geq 1}\frac{1}{mn(m+n)}=\color{red}{\frac{\zeta(3)}{8\pi^2}}$$
$$ \sum_{c\geq 1}\sum_{a\geq 1}\sum_{b\geq 1}\int_{0}^{+\infty}\frac{t^2(t^2-4c^2\pi^2)\,dt}{(t^2+4\pi^2 a^2)(t^2+4\pi^2 b^2)(t^2+4\pi^2 c^2)^2}\\=\frac{1}{16\pi^2}\sum_{c,a,b\geq 1}\frac{ab-c^2}{(a+b)(a+c)^2(b+c)^2}.\tag{3}$$
The first series is well-known, for instance:
$$\sum_{m,n\geq 1}\frac{1}{mn(m+n)}=\sum_{n\geq 1}\frac{2H_{n-1}}{n^2}=2\int_{0}^{1}\frac{\log(1-x)\log x}{x}\,dx = \int_{0}^{1}\frac{\log^2 x}{1-x^2}\,dx.$$
What about the second one? It can be written as an integral over the unit cube, so I believe it can be computed through the techniques outlined by C. Viola in Birational transformations and values of the Riemann zeta-function, paragraph $4$, Permutation groups for triple integrals. By exploiting the $a\leftrightarrow b$ symmetry we have:
$$\begin{eqnarray*} \sum_{c,a,b\geq 1}\frac{ab-c^2}{(a+b)(a+c)^2(b+c)^2} &=& \sum_{c,a,b\geq 1}\frac{(a+c)(b-c)}{(a+b)(a+c)^2(b+c)^2}\\&=&\sum_{c,a,b\geq 1}\left(\frac{1}{a+c}-\frac{1}{a+b}\right)\frac{1}{(b+c)^2}\end{eqnarray*} $$
suggesting that the last series in $(3)$ is just zero, but there are some convergence issues: we are allowed to exchange $a$ and $b$, but we are not allowed to exchange $b$ and $c$. We have:
$$ \sum_{a,b\geq 1}\left(\frac{1}{a+c}-\frac{1}{a+b}\right)\frac{1}{(b+c)^2}=\frac{1}{c^3}+\frac{\gamma+\psi(c)}{c^2}-\frac{\psi'(c)}{c}-\frac{1}{2}\psi''(c)\tag{4}$$
hence:
$$ \sum_{c\geq 1}\sum_{a,b\geq 1}\frac{ab-c^2}{(a+b)(a+c)^2(b+c)^2}=-\frac{1}{2}\sum_{c\geq 1}\psi''(c)\tag{5}$$
and:
$$ I = \color{red}{\frac{\zeta(3)}{2\pi^2}-\frac{1}{24}} = 0.01923024745\ldots \tag{6}$$
The last step follows from the integral representation for the $\psi$ function:
$$\sum_{c\geq 1}\psi''(c)=-\sum_{s\geq 0}\int_{0}^{1}\frac{x^s\,\log^2 x}{1-x}\,dx = -\int_{0}^{1}\frac{\log^2 x}{(1-x)^2}\,dx = -2\zeta(2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
system of modular equations. $x\equiv 2\pmod3$
$x\equiv 3\pmod 5$
$x\equiv 7 \pmod{11}$
How can I solve this system for $x$? I've tried all kinds of things using divisibility but no success. Any hints of solutions are greatly appreciated. What value of $x$ satisfies these three equations?
| Here is a hands-on solution:
$x\equiv 2\pmod3$ implies $x=2+3y$.
$x\equiv 3\pmod 5$ then implies $3y\equiv 1 \pmod 5$ and so $y\equiv 2 \pmod 5$ because $2 \cdot 3 \equiv 1 \pmod 5$. This means that $y=2+5z$ and $x=2+6+15z=8+15z$.
$x\equiv 7 \pmod{11}$ then implies $15z \equiv -1 \pmod{11}$ and so $z \equiv -3 \pmod{11}$ because $15 \equiv 4 \pmod{11}$ and $3 \cdot 4 \equiv 1 \pmod{11}$. This means that $z=-3+11t$ and so $x=8-45+165t=-37+165t$.
Thus, the solution is $x \equiv -37 \equiv 128 \pmod{165}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$\lim\limits_{x\rightarrow 0 }\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{ n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}$ From the post Evaluating limit (iterated sine function) and some discussions inside, one can collect the
following three limits
\begin{eqnarray*}
\lim_{x\rightarrow 0 }\frac{\sin _{n}x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{x-\sin _{n}x}{x^{3}} &=&\frac{n}{6} \\
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}}{x^{5}} &=&%
\frac{n^{2}}{24}-\frac{n}{30}.
\end{eqnarray*}
where $\sin _{n}x=\sin (\sin \cdots (\sin x)),\ n$ times composition.
So the next question would be, what is the following limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin _{n}x-x+\frac{n}{6}x^{3}-\left( \frac{%
n^{2}}{24}-\frac{n}{30}\right) x^{5}}{x^{7}}
\end{equation*}
and what are those corresponding limits after order 7 $?$
| Direct computation
$\def\wi{\subseteq}$
$\sin(x) \in x - \frac{1}{6} x^3 + \frac{1}{120} x^5 - \frac{1}{5040} x^7 + O(x^9)$ as $x \to 0$.
As $x \to 0$ and given function $f$ such that $f(x) \in x + a x^3 + b x^5 + c x^7 + O(x^9)$:
$\sin(f(x))$
$\ \in \left( x + a x^3 + b x^5 + c x^7 + O(x^9) \right)$
$\hphantom{\ \in} - \frac{1}{6} \left( x + a x^3 + b x^5 + O(x^7) \right)^3$
$\hphantom{\ \in} + \frac{1}{120} \left( x + a x^3 + O(x^5) \right)^5$
$\hphantom{\ \in} - \frac{1}{5040} \left( x + O(x^3) \right)^7$
$\hphantom{\ \in} + O(x^9)$
$\ \wi x + (a-\frac{1}{6}) x^3 + (b-\frac{1}{2}a+\frac{1}{120}) x^5 + (c-\frac{1}{2}b-\frac{1}{2}a^2+\frac{1}{24}a^2-\frac{1}{5040}) x^7$
[Thus we can set up recurrences for $a,b,c$ on iteration of applying $\sin$ and solve them to finish.]
Notes
As you can see, we only need to handle the coefficient sequences. I'm too lazy to figure out the computational complexity of this algorithm in general but it is quite bad because the numbers grow exponentially. Is there a faster method? I doubt so. No wonder Claude got tired! But at least this is implementable on a computer and we don't have to do it ourselves for high order approximations.
| {
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"url": "https://math.stackexchange.com/questions/1346667",
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"source": "stackexchange",
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"answer_count": 2,
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Proof the series is finite using following inequality Let
$$a_n=\frac1{\sqrt1}+\frac1{\sqrt 2}+\ldots +\frac1{\sqrt n}-2\sqrt n $$
For the task to prove that $$\tag1-2\le a_n\le -1 $$
I was given the hint
$$\tag2\sqrt{k+1}-\sqrt k<\frac1{2\sqrt k},\qquad \forall k\in\mathbb N^* $$
I managed to prove $(1)$ by other methods but woonder how to proof it actually using $(2)$.
| First, let us prove (2), that is
\begin{equation*}
\sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}},\ for\ k\geq 1.
\end{equation*}
\begin{eqnarray*}
\sqrt{k+1}-\sqrt{k} &=&\frac{\left( \sqrt{k+1}-\sqrt{k}\right) \left( \sqrt{%
k+1}+\sqrt{k}\right) }{\left( \sqrt{k+1}+\sqrt{k}\right) } \\
&=&\frac{1}{\left( \sqrt{k+1}+\sqrt{k}\right) } \\
&\leq &\frac{1}{\left( \sqrt{k}+\sqrt{k}\right) } \\
&=&\frac{1}{2\sqrt{k}}.
\end{eqnarray*}
Now let us prove the inequality
\begin{equation*}
-2\leq a_{n}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}%
-2\sqrt{n}.
\end{equation*}
Consider the inequality already proved, for $k=1,$ 2, $\cdots ,(n-1),$ and
write then vertically
\begin{eqnarray*}
\sqrt{2}-\sqrt{1} &\leq &\frac{1}{2\sqrt{1}} \\
\sqrt{3}-\sqrt{2} &\leq &\frac{1}{2\sqrt{2}} \\
&&\vdots \\
\sqrt{n}-\sqrt{n-1} &\leq &\frac{1}{2\sqrt{n-1}}
\end{eqnarray*}
Now add them all, they telescope (!), that is $\sqrt{2}$ of the first line
cancel with $\sqrt{2}$ of the second line, and so on, until $\sqrt{n-1}$ of
the (n-2)$^{th}$ line which cancel with $\sqrt{n-1}$ of the last line, to get%
\begin{equation*}
\sqrt{n}-\sqrt{1}\leq \frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+\cdots +\frac{1%
}{2\sqrt{n-1}}
\end{equation*}
then
\begin{equation*}
\sqrt{n}-1\leq \frac{1}{2}\left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}%
+\cdots +\frac{1}{\sqrt{n-1}}\right)
\end{equation*}
then
\begin{equation*}
2\sqrt{n}-2\leq \left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1%
}{\sqrt{n-1}}\right)
\end{equation*}
which implies
\begin{equation*}
-2\leq \left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{%
n-1}}\right) -2\sqrt{n}=a_{n}.
\end{equation*}
As was to be shown.
It remains to prove the right inequality.
${\bf UPDATE}$ To prove that
\begin{equation*}
a_{n}\leq -1,\ for\ all\ n\geq 1.
\end{equation*}
I was not able to prove it using the inequality
\begin{equation*}
\sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}},\ for\ k\geq 1.
\end{equation*}
However, if we compute
\begin{equation*}
a_{n+1}-a_{n}=\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}\left( \sqrt{n+1}+\sqrt{n}%
\right) }<0.
\end{equation*}
So the sequence (a$_{n})$ is decreasing, and since $a_{1}=\frac{1}{\sqrt{1}}%
-2\sqrt{1}=-1$ then
\begin{equation*}
a_{n}\leq a_{1}=-1,\ \ for\ all\ n\geq 1.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the remainder when the sum is divided by $1000$
Find $S \pmod{1000}$ given: $$S = \sum_{n=0}^{2015} n! + n^3 - n^2 + n - 1$$
$$S_0 = 0! + 0 - 0 + 0 -1 = 0$$
$$S_1 = 1! + 1 - 1 + 1 - 1 = 1$$
$$S_2 = 2! + 8 - 4 + 2 - 1 = 7$$
This isn't helping, so:
$n! = n(n-1)(n-2)...(1)$ but that is too complicated.
The $n!$ term is the hardest one. Besides that:
$$S = \sum_{n=0}^{2015} n! + \frac{2015^2 2016^2}{4} - \frac{(2015)(2016)(4031)}{6} + \frac{2015(2016)}{2} - 2015$$
$$\equiv \sum_{n=0}^{2015} n! + 25(4) - 15(336)(31) + 15(8) - 15 \pmod{1000}$$
Please offer hints, thank you!
| HINT :
Since $1000=2^3\cdot 5^3$, note that
$$15!\equiv 0\pmod{1000}.$$
Hence, we have
$$\sum_{n=0}^{2015}n!\equiv \sum_{n=0}^{\color{red}{14}}n!\pmod{1000}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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floor function problems There are two parts of my problem.
*
*Given $n$ and $x$, $\lfloor \frac nx \rfloor = q$, what is the maximum
possible value for $x$ such that we obtain the same floor value? i.e.
$\lfloor \frac nx \rfloor = \lfloor \frac{n}{x'}\rfloor = q$, where $x' \geq x $.
I think $x' = \lfloor \frac nq \rfloor$ is the answer.
*How many possible values(distinct) for q are there for all $x$ where $1\leq x\leq n$?
My thinking was that, what is the maximum value of $q$ that we can obtain
and also the minimum value of $q$. Then, the answer would be $max - min + 1$. But this is surely wrong because we can have same floor values for different values of x. For example, for any $x \geq \lfloor \frac n2\rfloor + 1$, $q = \lfloor \frac nx\rfloor = 1$. Thus the maximum number of distinct floor values is at most $1 + \lfloor \frac n2 \rfloor$. How do I reduce the upper bound since I have read that there are at most $2\sqrt n$ distinct vaues?
| By definition of $\lfloor\cdot\rfloor$, we have
$$\biggl\lfloor \frac{n}{x}\biggr\rfloor = q \iff q \leqslant \frac{n}{x} < q+1$$
for $q\in \mathbb{Z}$. Since we are here dealing with positive integers, the sense of the inequalities is retained when multiplying with $x$ and dividing by $q$ resp. $q+1$, so we have
$$\biggl\lfloor \frac{n}{x}\biggr\rfloor = q \iff \frac{n}{q+1} < x \leqslant \frac{n}{q}.$$
Using that also $x$ shall be an integer, we obtain
$$\biggl\lfloor \frac{n}{x}\biggr\rfloor = q \iff \biggl\lfloor \frac{n}{q+1}\biggr\rfloor < x \leqslant \biggl\lfloor\frac{n}{q}\biggr\rfloor.$$
Hence, for each $q$ there are $\bigl\lfloor \frac{n}{q}\bigr\rfloor - \bigl\lfloor \frac{n}{q+1}\bigr\rfloor$ integer values $1 \leqslant x \leqslant n$ so that $\bigl\lfloor \frac{n}{x}\bigr\rfloor = q$, and the maximal such $x$ is indeed $\bigl\lfloor \frac{n}{q}\bigr\rfloor$.
How do I reduce the upper bound since I have read that there are at most $2\sqrt{n}$ distinct values?
Note that for $x > \sqrt{n}$, we have $\frac{n}{x} < \sqrt{n}$ and hence $\bigl\lfloor \frac{n}{x}\bigr\rfloor \leqslant \lfloor \sqrt{n}\rfloor$. Thus we have at most $\lfloor \sqrt{n}\rfloor$ distinct values of $\bigl\lfloor \frac{n}{x}\bigr\rfloor$ for $1 \leqslant x \leqslant \lfloor \sqrt{n}\rfloor$, and the $\lfloor \sqrt{n}\rfloor$ values $1 \leqslant q \leqslant \lfloor \sqrt{n}\rfloor$ for $\lfloor \sqrt{n}\rfloor + 1 \leqslant x \leqslant n$, which gives at most $2\lfloor \sqrt{n}\rfloor$ distinct values total. In fact, the values of $\bigl\lfloor \frac{n}{x}\bigr\rfloor$ for $1 \leqslant x \leqslant \lfloor \sqrt{n}\rfloor$ are all distinct, and all values $1 \leqslant q \leqslant \lfloor \sqrt{n}\rfloor$ are attained, so there are exactly $2\lfloor \sqrt{n}\rfloor$ distinct values, unless the value $\lfloor \sqrt{n}\rfloor$ appears in both sets - when $\bigl\lfloor \frac{n}{\lfloor \sqrt{n}\rfloor}\bigr\rfloor = \lfloor \sqrt{n}\rfloor$, or equivalently $n < \lfloor \sqrt{n}\rfloor\cdot (\lfloor \sqrt{n}\rfloor + 1)$ - in which case there are exactly $2\lfloor \sqrt{n}\rfloor - 1$ distinct values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix inverses - Why are they derived the way they are? Note that this is not a question of how, but why. I know the mechanics of it, but this is the first thing i've come across that truly seems like magic, rather than a rigorous mathematical process.
There are questions on SE about proofs for the inverse, but niether here nor anywhere else on the internet, can I find a decent explanation of why it works the way it does:
So, basically, suppose the matrix is $\begin{bmatrix} a && b \\ c && d \end{bmatrix}$
Why do we swap values $a$ and $d$?
Why do $c$ and $b$ become negative?.
Why do we divide $a$, $b$, $c$ and $d$ by $ad-bc$?
Once again, this truly seems like magic. It's like ''hey, swap these fellers around, twiddle these signs a little, and poof, you're inverse!''.
Thanks in advance for any help :)
| You can derive it by simultaneous transforming $(A|I)\to(I|A^{-1})$. If $a \ne 0$ and $ad-bc\ne 0$ we get:
$$
\left[
\begin{array}{rr|rr}
a & b & 1 & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a & 1/a & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a & 1/a & 0 \\
0 & d - cb/a & -c/a & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a & 1/a & 0 \\
0 & (ad - bc)/a & -c/a & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & b/a & 1/a & 0 \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & 1/a + bc/a(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & ((ad-bc)+bc)/a(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & d/(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
$$
If $a = 0$ and $ad - bc \ne 0$ we have $b\ne 0$ and $c\ne 0$ and it goes like this:
$$
\left[
\begin{array}{rr|rr}
0 & b & 1 & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
c & d & 0 & 1 \\
0 & b & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & d/c & 0 & 1/c \\
0 & 1 & 1/b & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & -d/cb & 1/c \\
0 & 1 & 1/b & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & d/(-bc) & -b/(-bc) \\
0 & 1 & -c/(-bc) & 0/(-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & d/(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & 0/(ad-bc)
\end{array}
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Find the limit of $(2\sin x-\sin 2x)/(x-\sin x)$ as $x\to 0$ without L'Hôpital's rule I wonder how to do this in different way from L'Hôpital's rule:
$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$
Please help me solve this without using L'Hopital's rule.
| Write
\begin{eqnarray*}
\frac{2\sin x-\sin 2x}{x-\sin x} &=&\frac{2\sin x-2\sin x\cos x}{x-\sin x} \\
&=&2\left( \frac{\sin x}{x}\right) \left( \frac{1-\cos x}{x^{2}}\right)
\left( \frac{x^{3}}{x-\sin x}\right)
\end{eqnarray*}
Using standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6}
\end{eqnarray*}
it follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x-\sin x} &=&2\lim_{x\rightarrow
0}\left( \frac{\sin x}{x}\right) \lim_{x\rightarrow 0}\left( \frac{1-\cos x}{%
x^{2}}\right) \lim_{x\rightarrow 0}\left( \frac{x^{3}}{x-\sin x}\right) \\
&=&2\left( 1\right) \left( \frac{1}{2}\right) \left( \frac{6}{1}\right) \\
&=&6.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(bx+d)^2$
Thanks for any help.
| Hint We can write the given cubic polynomial as
$$ax^3 + bx^2 + cx + d = a(x - \alpha) (x - \beta) (x - \gamma).$$
Expanding the r.h.s. and comparing like terms gives
\begin{align}
b &= -a(\alpha + \beta + \gamma) \\
&\,\,\vdots
\end{align}
(These are essentially Vieta's Formulas.)
Additional hint Similarly, the cubic polynomial $A X^3 + B X^2 + C X + D$ whose roots are $\alpha^2, \beta^2, \gamma^2$ can be written as $$A X^3 + B X^2 + C X + D = (x - \alpha^2) (x - \beta^2) (x - \gamma^2),$$ and again expanding the r.h.s. gives, e.g., \begin{align} B &= -(\alpha^2 + \beta^2 + \gamma^2) \\ &\,\,\vdots \end{align} (Of course, this polynomial is only determined up to multiplication by a nonzero constant, which in particular does not change the roots. The choice of $1$ for the leading coefficient above simplifies some computations.) All that remains now is to write the coefficients $B, C, D$ in terms of the coefficients $a, b, c, d$. The two applications of Vieta's Formulas above expressions all of these coefficients in terms of symmetric functions of the roots $\alpha, \beta, \gamma$, so the coefficients can be related in terms of Newton's Identities, which give, e.g., $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\beta \gamma + \gamma \alpha + \alpha \beta).$$ Rearranging earlier equations and substituting give e.g., $$-B = \left(-\frac{b}{a}\right)^2 - 2 \left(\frac{c}{a}\right),$$ or $$B = \frac{2 a c - b^2}{a^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Straight line is tangent to the curve. The straight line $y=mx+1$ is tangent to the curve $x^2+y^2-2x+4y=0$. Find the possible values of $m$.
My attempt
Substitute the $y=mx+1$ into the equation $x^2+y^2-2x+4y=0$. $$x^2+(mx+1)^2-2x+4(mx+1)=0$$ $$x^2+m^2x^2+2mx+1-2x+4mx+4=0$$ $$(1+m^2)x^2+6mx-2x+5=0$$ $$(1+m^2)x^2+(6m-2)x+5=0$$
I think what I did is wrong as I don't know how to continue from my steps. Can anyone explains it? Thanks
| If line $y=mx+1$ is a tangent to $F(x, y)=0$, where $F(x, y)$ is a polynom of degree $2$, then $F(x, mx+1)=0$ have exactly one solution. Hence, discriminant is zero: $$(6m-2)^2=4\cdot(1+m^2)\cdot5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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a sequence limit with inequality condition Let sequence $\{a_{n}\}$ such
$$\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}},n\ge 1$$
Find limits
$$\lim_{n\to\infty}n\left(\dfrac{9}{16}-a_{n}\right)$$
I am working on a problem and I am lead to prove the following inequality
$$\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}\ge 2\sqrt{n}$$
and
$$\sqrt{na_{n}+n}+\sqrt{na_{n}}\le 2\sqrt{n}$$
| We have $\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1}\le\dfrac{\sqrt{n}}{2}\le\sqrt{na_{n}+n}-\sqrt{na_{n}}$ so
$(\sqrt{na_{n}+n+1}-\sqrt{na_{n}+1})\dfrac{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}\le\dfrac{\sqrt{n}}{2}\le(\sqrt{na_{n}+n}-\sqrt{na_{n}})\dfrac{\sqrt{na_{n}+n}+\sqrt{na_{n}}}{\sqrt{na_{n}+n}+\sqrt{na_{n}}}$
giving $\dfrac{n}{\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}}\le\dfrac{\sqrt{n}}{2}\le\dfrac{n}{\sqrt{na_{n}+n}+\sqrt{na_{n}}}$.
Then $\sqrt{na_{n}+n+1}+\sqrt{na_{n}+1}\ge 2\sqrt n\ge\sqrt{na_n+n}+\sqrt{na_n}$.
It comes $na_n+n+1\ge 4n+na_n+1-4\sqrt n\sqrt{na_{n}+1}$ or else $4\sqrt n\sqrt{na_{n}+1}\ge 3n$ and $4n-4\sqrt n\sqrt{na_n}+na_n\ge na_n+n$ or else $3n\ge 4\sqrt n\sqrt{na_n}$.
Thus $a_n\leq\dfrac{9}{16}\leq a_n+\frac{1}{n}$.
Now if $a_n=\dfrac{9}{16}$, the sought limit is 0 but if $a_n=\frac{9}{16}-\frac{1}{n}$, the limit is 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.
Attempt: Denoting partial sums by $S_n$, we have
\begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1-x^{n+1}).
\end{eqnarray*}
I tried to show uniform convergence by showing convergence of $\sup_{x \in [0,1]} S_n(x)$, but I seem unable to solve the resulting first order condition to find the maximizing x. Is there a simpler way to do this?
If I could show uniform convergence, I could integrate the infinite series term by term yielding
\begin{eqnarray*}
\int \frac{(1-x^2)^2}{1-x} dx &=& \sum_{n=0}^\infty \int_0^1 (1-x^2)^2 x^n dx \\ &=&\sum_{n=0}^\infty \int_0^1 x^n - 2x^{2+n} + x^{4+n} dx
\\ &=& \dots
\\ &=&\sum_{n=0}^\infty \frac{8}{(n+1)(n+3)(n+5)}
\\ &=&\sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}
\end{eqnarray*}
and I would be done.
| \begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1).
\end{eqnarray*}
the uniform convergence follows from
\begin{eqnarray*}
S_m -S_n&=& \sum_{k=n+1}^m (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x}\xrightarrow[n \to \infty]{} 0 \\
\end{eqnarray*}
Since $x^{m+1} \leq x^{n+1} \to 0$
Edit: To get uniform convergence one needs further to note that $x^{m+1},x^{n+1} \leq 1$ and that $1-x^2 = 1+x (1-x) $ so $$(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x} = (1+x)^2(1-x)x^{n+1}-x^{m+1}.$$
So for $x>1-\delta$
$$(1+x)^2(1-x)x^{n+1}-x^{m+1}\leq 2\delta $$
and for $x<1-\delta$ $x^n \leq (1-\delta)^n \to 0$
$$(1+x)^2(1-x)x^{n+1}-x^{m+1} \leq 2 (1-\delta)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verify if 2 functions are inverse to each other According this site, 2 functions $f$ and $g$ are the inverse function of each other, only if both $(f \circ g) (x) = x$ and $(g \circ f) (x) = x$ are true.
Is it really necessary to prove both of them? Can someone please provide an counter example where one is satisfied and the other is not?
Edit: what if given $f$ and $g$ have the same domain?
Edit2: what if $f$ and $g$ are both monotonic?
| Try $f(x) = x^2$ and $g(x) = \sqrt{x}$. Then $(f \circ g)(x) = x$, but $(g \circ f)(-1) \neq -1$.
EDIT: If you want $f$ and $g$ to have the same domain, let
$$f(x) = \left\lbrace \begin{array}{rcl}(x + 1)^3 & : & x \le -1 \\ (x - 1)^3 & : & x \ge 1 \\ 0 & : & -1 < x < 1 \end{array} \right.$$
Notice that $f$ definitely is not invertible, since it isn't one-to-one. Also let
$$g(x) = \left\lbrace \begin{array}{rcl}\sqrt[3]{x}-1 & : & x < 0 \\ \sqrt[3]{x}+1 & : & x \ge 0 \end{array} \right.$$
Both functions have a domain of $\mathbb{R}$. Now, I claim that $(f \circ g)(x) = x$ for any $x$. We have two possibilities: $x \ge 0$ and $x < 0$. If $x \ge 0$, then $g(x) = \sqrt[3]{x} + 1 \ge 1$, so
$$(f \circ g)(x) = f(\sqrt[3]{x} + 1) = (\sqrt[3]{x} + 1 - 1)^3 = x.$$
On the other hand, if $x < 0$, then $g(x) = \sqrt[3]{x} - 1 < -1$, so
$$(f \circ g)(x) = f(\sqrt[3]{x} - 1) = (\sqrt[3]{x} - 1 + 1)^3 = x.$$
On the other hand, we cannot have $(g \circ f)(x) = x$, otherwise $f$ would be invertible and hence injective. Specifically, we can compute
$$(g \circ f)(1/2) = g(0) = 1 \neq 1/2.$$
EDIT2: Free bonus! Both $f$ and $g$ are monotonic!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:
Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.
I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{\sqrt{2}} \\[4pt]
\left|\sin\theta + \cosθ\right| &< \frac{1}{\sqrt{2}} \\[4pt]
\left(\sin\theta + \cos\theta\right)^2 &< \frac12 \\[4pt]
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta &< \frac12 \\[4pt]
1 + 2\sin\theta\cos\theta &< \frac12 \\[4pt]
1 + \sin 2\theta &< \frac12 \\[4pt]
\sin 2\theta &< -\frac12
\end{align}$$
But I don't know where to go from here. Can someone help me figure out how to get to the answer in the book: $105^\circ < θ < 165^\circ$. Thanks!
| To solve: $$\sin(2\theta) < -\frac{1}{2}$$
we treat $2\theta$ as a single variable. The domain in your problem is $ 0^\circ < \theta < 180^\circ$. So to get our new domain all we must do is multiply by $2$ to obtain: $0^\circ < 2\theta < 360^\circ$.
Now we must ask ourselves what angles when put through the sine function give me $-\frac{1}{2}$? Thinking back to the unit circle we have that $$\sin(210^\circ) = -\frac{1}{2} = \sin(330^\circ)$$
So any values between these (again from the unit circle) give me a value less than $-\frac{1}{2}$.
$$210^\circ < 2\theta < 330^\circ$$
Dividing by $2$ we get the result:
$$105^\circ < \theta < 165^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2}
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$?
| Dividing the numerator by $x\sqrt x\sqrt x$ and denominator by $x^2$, we have
$$ \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
= \lim_{x \to \infty} \frac{\sqrt{1+\dfrac1x}\left(\dfrac1{\sqrt x}-\sqrt{2+\dfrac3{\sqrt x}}\right)}{\dfrac7{x^2}-\dfrac6x+4}=\frac{1\cdot(0-\sqrt2)}4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Solve this exponential equation: $3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$ I tried solving this equation $$3^{2x}+\left(\frac{1}{2}\right)^{-x} \cdot 3^{x+1}-2^{2x+2}=0$$ by taking the log of both sides, but with no results, what do I do? Sorry if this equation is very easy, I couldn't solve it...
| Hint: This reduces to $$3^{2x} + 3\cdot 2^x \cdot 3^x - 4\cdot 2^{2x} = 0$$
If $a=3^x$ and $b=2^x$ then we get $$a^2 + 3ab - 4b^2 = 0.$$
More of a hint: You have $x = \log_3 a = \log_2 b$. So now you have two equations involving $a$ and $b$...
| {
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"url": "https://math.stackexchange.com/questions/1364578",
"timestamp": "2023-03-29T00:00:00",
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Proof question: Prove that 2^(odd integer) + 5^(odd integer) + 2 is a multiple of 3, and 4^(any integer) + 1 can be expressed as 5n, 5n + 1 or 5n + 2. I had been working on the claim in the above question for sometime now. Statistically speaking, it works. For example:
$$\left(2^1 \right)+\left(5^1\right)+2=9,\\
\left(2^3\right)+\left(5^1\right)+2=15,\\ \ldots$$
It works with all the integers that I tested with. Can we give a simple proof for this one?
The second part is like this:
$$\left(4^1\right)+1=5,$$
$$\left(4^2\right)+1=17 = 15+2,$$
$$\left(4^3\right)+1=65,$$
$$\left(4^4\right)+1=257 = 255+2,\\ \ldots$$
So, the numbers are all expressed as $5n$, $5n+1$ or $5n+2$. Is there a generalised proof for this observation?
Thanks.
| For questions of divisibility, it helps to think in terms of modular arithmetic.
To show something is a multiple of $3$, we can check to see if it is $0 \pmod 3$:
$$2^{2m+1} + 5^{2n+1} + 2 \equiv -1 + 2 + 2 \pmod 3 \equiv 3 \pmod 3 \equiv 0 \pmod 3$$
Note: $5$ raised to any power always ends in $5$, and $5 \equiv 2 \pmod 3$. As well, $2 \equiv -1 \pmod 3$ and $-1$ raised to an odd power is always $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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simplify and evaluate $\frac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$ How do you simplify and evaluate $\dfrac{\tan80^\circ-\tan20^\circ}{1+\tan80^\circ\tan20^\circ}$?
What is the problem asking?
| Method 1: Notice formula, $$\color{red}{\frac{\tan A-\tan B}{1+\tan A\tan B}=\tan (A-B)}$$ Hence, we have $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$$$=\tan (80^\circ-20^\circ)$$ $$ =\tan (60^\circ)$$ $$ =\color{blue}{\sqrt{3}}$$
Method 2: Notice, $$\color{red}{\sin A\cos B-\cos A\sin B=\sin (A-B)}$$ & $$\color{red}{\cos A\cos B+\sin A\sin B=\cos (A-B)}$$ Now, we have $$\frac{\tan 80^\circ-\tan 20^\circ}{1+\tan 80^\circ\tan 20^\circ}$$ $$=\frac{\frac{\sin 80^\circ}{\cos 80^\circ}-\frac{\sin 20^\circ}{\cos 20^\circ}}{1+\frac{\sin 80^\circ}{\cos 80^\circ}\frac{\sin 20^\circ}{\cos 20^\circ}}$$ $$=\frac{\sin 80^\circ\cos 20^\circ-\cos 20^\circ\sin 20^\circ}{\cos 80^\circ\cos 20^\circ+\sin 80^\circ\sin 20^\circ}$$ $$=\frac{\sin(80^\circ-20^\circ)}{\cos(80^\circ-20^\circ)}$$$$=\frac{\sin 60^\circ}{\cos 60^\circ}$$ $$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$ $$ =\color{blue}{\sqrt{3}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\
2\theta = 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\
\theta = k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
| $$z^3+z^*=0.$$
Let $z=x+iy$. Then
$$(x+iy)^3+x-iy=0,$$ so that
$$x+x^3-3xy^2=0,\tag{1}$$ and $$-y+3x^2y-y^3=0.\tag{2}$$
We ignore the trivial solution $x=y=0$ and suppose that $x,y\neq 0$. Then using (1) and (2) we see that $$x^3+x=x\left(1+\frac{y^3+y}{3y}\right).$$ Hence, $$-3xy^2+x\left(1+\frac{y^3+y}{3y}\right)=0.$$ Since $x\neq 0$ this gives $$-3y^2+1+\frac{y^3+y}{3y}=0,$$ leading to $$y(1-2y^2)=0.$$ Therefore, we must have $$y=\pm\frac{1}{\sqrt{2}}.$$ Substituting into $(2)$ gives $$x=\pm\frac{1}{\sqrt{2}}.$$ Therefore,
$$z=x+iy=\pm\frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}},$$ where each $\pm$ is chosen independently of the other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Get amount of submatrixes from $a \times b $matrix I was trying to do the following exercise
Given a grid of size $a \times b$, write a formula able t calculate the total number of rectangles contained in this rectangle. All integer sizes and positions are counted.
Examples:
$3 \times 2$ matrix $\Rightarrow 18$
$4 \times 4$ matrix $\Rightarrow 100$
Here is how the $3\times2$ grid works:
$1$ rectangle of size $3\times 2$:
[][][]
[][][]
$2$ rectangles of size $3\times 1$:
[][][]
$4$ rectangles of size $2\times 1$:
[][]
$2$ rectangles of size $2\times 2$
[][]
[][]
$3$ rectangles of size $1\times 2$:
[]
[]
$6$ rectangles of size $1\times 1$:
[]
As you can see $(1 + 2 + 4 + 2 + 3 + 6) = 18$, and is the solution for the $3\times 2$ grid.
...
Alledgedly, the answer is $\frac{((ab)+a)((ab)+b))}{4}$
I can't understand why, can anybody explain this?
| Yes, although I find it easier to understand it as $\frac{a(a+1)b(b+1)}{4}$.
Every rectangle can be described by selecting two not necessarily distinct columns (the columns the corners are going to occupy) and two not necessarily distinct rows (the rows the corners are going to occupy).
There are $\binom{a}{2}=\frac{a(a-1)}{2}$ ways to select distinct columns and $a$ ways to select only one column. So $\frac{a(a+1)}{2}$ ways in total.
There are $\binom{b}{2}=\frac{b(b-1)}{2}$ ways to select distinct rows and $b$ ways ways to select only one row, so $\frac{b(b+1)}{2}$ in total.
So there are $\frac{a(a+1)}{2}\frac{b(b+1)}{2}=\frac{a(a+1)b(b+1)}{4}=\frac{(ab+a)(ab+b)}{4}$ ways to select the rows and columns of the corners of the rectangle, this uniquely determines each rectangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x^2 + (k-3)x + k = 0$, ranges of k for roots to be of same sign I need some help on the following.
The quadratic that I am dealing with is $x^2 + (k-3)x + k = 0$, and I need to find ranges of values of $k$, for which the roots will have the same sign.
For the roots of $ax^2+bx+c=0$ to have same signs, $a(x^2+ \frac{b}{a}x+\frac{c}{a})$, the last term, i.e. $\frac{c}{a} > 0$, because if you factorize the quadratic, to arrive at positive constant you either have to have two negative numbers multiplied or two positive multiplied by each other.
I will show what I have done.
I first thought that the roots will have same sign if they are the same, i.e. $b^2-4ac=0$, that led me to the result that $k=1$ or $k=9$.
Then I looked at $k$, $k>0$, according to what I said.
And the answer is actually $0<k<1$ and $k>9$. I have noticed that for this quadratic to have real and distinct roots, it has to satisfy following $k<1$ or $k>9$. Please help me to arrive at the required result. Thank you
| We solve the equation: \begin{align} x^2+(k-3)x+k = 0 &\implies x = \frac{3-k \pm \sqrt{k^2-6k+9 - 4\cdot 1\cdot k}}{2} \\ &\implies x = \frac{3-k \pm \sqrt{k^2-10k + 9}}{2}\end{align} For starters, we must have $k^2-10k + 9 = (k-1)(k-9) \geq 0$, which happens if $k \leq 1$ or $k \geq 9$. Recall that two numbers have the same sign if and only if their product is positive, and by Vieta's formulas, their product is $k$. So far, we have $0 < k \leq 1$ or $k \geq 9$. However, $k=1$ gives $1$ as a double root and $9$ gives $-3$ as a double root. So the book forgot these cases: our final answer is $0 < k \leq 1$ or $k \geq 9$.
| {
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$0<\int_0^\infty\frac{\sin t}{\ln(1+x+t)} dt<\frac{2}{\ln(1+x)}$ This is my first time posting so please excuse me if I don't follow the proper etiquette. This one is a rather hard problem that was assigned to me for my calculus 2 class. Thank you for your help!
For $x > 0$, prove that:
$$0<\int_0^\infty\frac{\sin t}{\ln(1+x+t)} dt<\frac{2}{\ln(1+x)}$$
| $$\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t = \int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + t + 2n\pi)} - \frac{1}{\ln(1 + x + t + (2n + 1)\pi)} \right)\right]$$
Note
$$f_n(t) = \frac{1}{\ln(1 + x + 2n\pi + t)} - \frac{1}{\ln(1 + x + (2n + 1)\pi + t)}.$$
The derivative
$$f_n'(t) = - \frac{1}{(1 + x + 2n\pi + t)\ln^2(1 + x + 2n\pi + t)} + \frac{1}{(1 + x + (2n + 1)\pi + t)\ln^2(1 + x + (2n + 1)\pi + t)} < 0.$$
Thus $\forall t > 0, \,\, f_n(t) < f_n(0) = \frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}$.
$$\begin{align}\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t &=\int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty f_n(t) \right] \\
&< \int_0^\pi \left[\sin(t)\mathrm{d}t \cdot \sum_{n=0}^\infty f_n(0) \right] \\
&= \left(\int_0^\pi \sin(t)\mathrm{d}t \right)\cdot \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right) \\
&= 2 \sum_{n=0}^\infty \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right)
\end{align}$$
Consider the partial sum :
$$\begin{align} S_{N} &= \sum_{n=0}^N \left(\frac{1}{\ln(1 + x + 2n\pi)} - \frac{1}{\ln(1 + x + (2n + 1)\pi)}\right) \\
&= \frac{1}{\ln(1 + x)} - \sum_{n=0}^{N-1} \left(\frac{1}{\ln(1 + x + (2n + 1)\pi)} - \frac{1}{\ln(1 + x + (2n + 2)\pi)}\right) - \frac{1}{\ln(1 + x + (2N + 1)\pi)} \\
&<\frac{1}{\ln(1 + x)}.
\end{align}$$
So
$$\int_0^\infty\frac{\sin t}{\ln(1+x+t)} \mathrm{d}t < \frac{2}{\ln(1 + x)}. $$
| {
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"url": "https://math.stackexchange.com/questions/1369512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate $\int \frac{\arctan x}{x^4} dx$? I have written the integral as $\int x^{-4} \arctan x dx$. Then, by applying by parts, I got $-3\dfrac{\arctan x}{x^3} + 3\int \dfrac{1}{x^3(1 + x^2)} dx$. Now, how can I solve the later integral? Is there any other trick to do this?
| \begin{eqnarray}
\int\frac{\arctan x}{x^4}\,dx&=&-\frac{\arctan x}{3x^3}+\frac13\int\frac{1}{x^3(1+x^2)}\,dx\stackrel{x=1/t}{=}-\frac{\arctan x}{3x^3}-\frac13\int\frac{1}{t^{-3}(1+t^{-2})t^2}\,dt\\[10pt]
&=&-\frac{\arctan x}{3x^3}-\frac13\int\frac{t^3}{1+t^2}\,dt=-\frac{\arctan x}{3x^3}-\frac13\int\frac{t(t^2+1)-t}{1+t^2}\,dt\\[10pt]
&=&-\frac{\arctan x}{3x^3}-\frac13\int\left(t-\frac{t}{1+t^2}\right)\,dt=-\frac{\arctan x}{3x^3}-\frac13\left[t^2-\frac12\ln(1+t^2)\right]+C\\[10pt]
&\stackrel{t=1/x}{=}&-\frac{\arctan x}{3x^3}-\frac13\left[\frac{1}{x^2}-\frac12\ln\left(1+\frac{1}{x^2}\right)\right]+C\\[10pt]
&=&-\frac13\left[\frac{\arctan x}{x^3}+\frac{1}{x^2}-\frac12\ln(1+x^2)+\ln|x|\right]+C
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction
Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$
As we have odd numbers that are squared we could use $n = 2k-1$.
But the $2$ sides do not equate for $n=1$ or $2k-1$ (if you set $k=1$).
Also need to find the sum to $2n+1$ terms.
|
As we have odd numbers that are squared we could use n = 2k-1
You are summing the sequence: $\{1^2-3^2, 5^2-7^2, 9^2-11^2, \ldots (4n-3)^2-(4n-1)^2, \ldots\}$
But the 2 sides do not equate for n=1 or 2k-1 (if you set k=1)
See above. The first term is $-8$, and so...
Also need to find the sum to (2n+1) terms.
You need to show that if the sum of the first $n$ terms is $-8 n^2$, then if you add $-16(n+1)+8$ you obtain $-8(n+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find the Jordan form and $P$ such that $P^{-1}AP = J$. Consider the matrix $$A = \left(\begin{array}{cccc} -11&0&-9\\32&1&24\\16&0&13 \end{array}\right)$$
I want to find the Jordan form of $A$, with $1$-s at the bottom and the jordan basis, which is $P$ columns such that $P^{-1}AP = J$.
I evaluated the charechteristic polynomial which is $f_A(x) = -(x-1)^3$. So we have one eigenvalue which is $\lambda =1$.
Now,
$$A-I = \left(\begin{array}{cccc} -12&0&-9\\32&0&24\\16&0&12 \end{array}\right)$$
$$\ker (A-I) = \text{span}\{(0,1,0),(1,0,3/4)\}$$
$$\ker (A-I)^2 =\mathbb{R}^3$$
As I understand I need to find a vector, $v_3$ such that $\ker (A-I) \cup v_3 = \ker (A-I)^2 = \mathbb{R}^3$.
Then, $$v_2 = (A-I)v_3 \\ v_1 = (A-I)v_2$$
And we (suppose to) get $P = (v_1,v_2,v_3)$ such that $P^{-1}AP = J$.
I chose $v_3 = (1,0,0)$, but it doesn't work, and for the matter of fact I'm confused by all this.
I'd be glad if you could explain to me how to correct this.
Thanks.
| Saying that 1 is an eigenvalue for matrix $A$ means that there exist non-zero vectors, $v$, such that $Av= v$ or $(A- I)v= 0$.
Here, that means $$\begin{bmatrix}-12 & 0 & -9 \\ 32 & 0 & 24 \\ 16 & 0 & 12\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\begin{bmatrix}-12x- 9z \\ 32x+ 24z \\ 16x+ 12z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$
You basically have that. Now note that this gives you the three equations $$-12x- 9z= 0,$$ $$32x+ 24z= 0,$$ and $$16x+ 12z= 0.$$ From the first equation, $9z= -12x$ so $z= (-12/9)x= (-4/3)x.$ Setting $z= (-4/3)x$ in the second and third equations gives $$32x+ 24(-4/3)x= 32x- 32x= 0$$ and $$16x+ 12z= 16x+ 12(-4/3)x= 16x- 16x= 0.$$
That is, for any value of $x, z= (-4/3)x$ will satisfy all those equations. Since $y$ did not enter into this, $y$ can be anything. That is, the eigenvectors must be of the form $$\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix} x \\ y \\ (-4/3)x \end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ -4/3\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}.$$
We can use $\begin{pmatrix}3\\ 0\\ -4\end{pmatrix}=3\cdot \begin{pmatrix}1\\ 0\\ -4/3\end{pmatrix}$ and $\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}$ form a basis for the Eigen space. since that is a two dimensional subspace, the "Jordan Normal Form" will be of the form $$J= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 0 \\0 & 0 & 3\end{bmatrix}$$ and we can form the matrix $$P^{-1}$$ using the vectors $$\begin{pmatrix}3 \\ 0\\ -4\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$$ as columns.
| {
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"url": "https://math.stackexchange.com/questions/1372634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| Not a visual proof, but by the Binomial Theorem, $$(1-x)^{-2}=\sum_0^{\infty}{-2\choose n}(-1)^nx^n$$ Now $${-2\choose n}={-2\cdot-3\cdots(-1-n)\over n!}=(-1)^n(n+1)$$ so $(1-x)^{-2}=\sum_0^{\infty}(n+1)x^n$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 13
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Trigonometric equation cos sin and power The problem is $2\cos t - 3\sin^2t +2 = 0$.
I get to $2\cos t -3\sin^2t =-2$
I think that I need to use a trigonometric identity like $\cos(x+y)$ and to divide $2\cos t -3\sin^2t$ with the $\sqrt{2^2+3^2}$
Do you know how to solve this? It should be $\sqrt{2^2 + 3^2}$
| We have, $$2\cos t-3\sin^2 t+2=0$$ $$\implies 2\cos t-3+3\cos^2 t+2=0$$ $$\implies 3\cos^2t+ 2\cos t-1=0$$ Factorizing the expression, we get $$(3\cos t-1)(\cos t+1)=0$$ $$\text{if}\ 3\cos t-1=0 \implies \cos t=\frac{1}{3}\implies \color{blue}{t=2n\pi\pm\cos^{-1}\left(\frac{1}{3}\right)}$$ $$\text{if}\ \cos t+1=0 \implies \cos t=-1 \implies \color{blue}{t=(2n+1)\pi}$$ Where, $n$ is any integer
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Checking logarithm inequality.
Which one of the following is true.
$(a.)\ \log_{17} 298=\log_{19} 375 \quad \quad \quad \quad (b.)\ \log_{17} 298<\log_{19} 375\\
(c.)\ \log_{17} 298>\log_{19} 375 \quad \quad \quad \quad (d.)\ \text{cannot be determined} $
$17^{2}=289 $ it has a difference of $9$ and $19^{2}=361$ it has a difference of $14$ .
I am not aware of any method if it is there to check such problems,
I would also prefer a method without calculus unless necessary.
I look for a short and simple way .
I have studied maths up to $12$th grade.
| Let $x=\log_{17}{298}, y=\log_{19}{375}$.
By definition of logarithms,
$17^x = 298$ and $19^y=375$
So
$17^{x-2} = \dfrac{298}{289} = 1 + \dfrac{9}{289} \tag{1}$ and $19^{y-2}=\dfrac{375}{361} = 1 + \dfrac{14}{361} \tag{2}$.
Now take natural logarithms
$(x-2)\ln{17} = \ln(1+\dfrac{9}{289}) \approx \dfrac{9}{289} \tag{3}$
and
$(y-2)\ln{19} = \ln(1+\dfrac{14}{361}) \approx \dfrac{14}{361} \tag{4}$
From $\ln{19} \approx \ln{17}(1+\frac{2}{17})$ and $\dfrac{14}{361} \times \dfrac{17}{19} \gg \dfrac{9}{289}$
we can say $\dfrac{\frac{14}{361}}{\ln{19}} > \dfrac{\frac{9}{289}}{\ln{17}}$
Then by equations (3), (4)
we have $y-2 > x-2$ or $\boxed{\log_{19}{375} > \log_{17}{298}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $2\sin^3x + \sin3x +3\sin^2x \cos x + \cos^3x=0$ $2\sin^3x + \sin3x +3\sin^2x\cos x + \cos^3x=0$
My try:
$$2\sin^3x +3\sin x - 4\sin^3x +\cos x(3\sin^2x+\cos^2x)=0 $$
$$ \cos x(2\sin^2x+1) - 2\sin^3x+3\sin x=0.$$
And then i have no idea.
| Since
$$
\sin 3x=3\cos^2x\sin x-\sin^3x
$$
you actually have
$$
\sin^3x+3\sin^2x\cos x+3\sin x\cos^2x+\cos^3 x=0,
$$
or, using the binomial theorem,
$$
(\sin x+\cos x)^3=0.
$$
I'm sure you can take it from here.
| {
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How to rewrite $\frac{d}{d(x+c)}$? I would like to know how to rewrite the following equations:
$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$
Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.
| $$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right) }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right) }{ dx } =0\\ \frac { df\left( x \right) }{ dx } \frac { 1 }{ \left( x+c \right) } -f\left( x \right) \frac { 1 }{ { \left( x+c \right) }^{ 2 } } =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand and right hand derivative using the concept of limits (That makes it huge)? Can intuition be used in any way?
| Expand $x-2\sqrt{2x-4}$ at $x=4$ to get $\frac{1}{8}(x-4)^2+O((x-4)^3)$. Hence
$$
\sqrt{x-2\sqrt{2x-4}} \approx \sqrt{\frac{1}{8}}|x-4|,
$$
and this should make it clear why $f$ is not differentiable at $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)\cdot
\left(x-2\sqrt{4-\sqrt{15}}+\sqrt{15}\right) \\
\cdot \left(x+2\sqrt{4-\sqrt{15}}+\sqrt{15}\right)
\cdot\left(x+2\sqrt{15 (4-\sqrt{15})}+8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)$$
| In general we would guess that $\sqrt a+\sqrt b+\sqrt c$ has eight conjugates, obtainable by toggling signs individually for the surds. However, in this special case we see that $\sqrt a\sqrt b\sqrt c=30$, which cannot change its sign. Hence once we picked the sign of two of the surds, the sign of the third is determined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Differential equation: $Ay'' + By' + Cy = h(x)$ I'm stuck solving the equation $y'' - 3y' + 2y = 2x^3-30$.
The auxiliary equation is $k^2 - 3k + 2 = 0$ where $k_1 = 1, k_2=3$. Thus the general solution is: $$y_g = C_1e^x + C_2e^{3x}$$
Then, I tried to find the particular solution taking into consideration that $h(x) = Q_n(x) \cdot e^{\alpha x}$ where $\alpha$ is zero and doesn't equal one of auxiliary equation's roots, and $n$ is the order of $h(x)$ and equals 3. So, I get the equations as:
$$ y_p = Ax^3 + Bx^2 +Cx + D, \\
y'_p = 3Ax^2 + 2Bx + C, \\
y''_p = 6Ax + 2B$$
Having substituted they in the initial equation I get the system:
$$ \begin{cases}
A = 2, \\
B + 3A = 0, \\
C + 2B + 6A = 0, \\
D + C + 2B = -30
\end{cases} \implies
\begin{cases}
A = 2, \\
B = -6, \\
C = 0, \\
D = -18
\end{cases}
$$
Thus my particular solution is $y_p = 2x^3 - 6x^2 - 18$ but the correct answer is: $$y_p = x^3 + \frac{9}{2}x^2 + \frac{21}{2}x - \frac{15}{4}$$
Where was I wrong?
| The equations you should have are: $2A=2 \\ -9A+2B=0 \\ 6A-6B+2C=0 \\ 2B-3C+2D=-30 \\ \text{ Solving this system should give you your solution. }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I solve this recurrence relation: $a_n = 3a_{n-1} + \frac{4^n}{4}$? How can I solve the following recurrence relation?
$$a_n = 3a_{n-1} + \frac{4^n}{4}$$
I know that $a_n^{(h)} = 3a_{n-1}$ and that the characteristic equation is:
$$r-3 = 0$$ and thus:
$$a_n^{(h)} = \alpha_1(3)^n$$
I am struggling with the particular solution $a_n^{(p)}$.
| Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write your recurrence as:
$$
a_{n + 1} = 3 a_n + 4^n
$$
Multiply by $z^n$, add over $n \ge 0$ to get:
$$
\sum_{n \ge 0} a_{n + 1} z^n
= 3 \sum_{n \ge 0} a_n z^n + \sum_{n \ge 0} 4^n z^n
$$
Recognize some sums:
$$
\frac{A(z) - a_0}{z}
= 3 A(z) + \frac{1}{1 - 4 z}
$$
Solve for $A(z)$, write as partial fractions:
$\begin{align}
A(z) &= \frac{a_0 - (4 a_0 - 1) z}{1 - 7 z + 12 z^2} \\
&= \frac{a_0 - 1}{1 - 3 z} + \frac{1}{1 - 4 z}
\end{align}$
This is just two geometric series:
$$a_n = (a_0 - 1) \cdot 3^n + 4^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
| Because $$(A-B)^{-1}\not=A^{-1}-B^{-1}.$$
Note that we have$$(\color{red}{A-B})^{-1}=\frac{1}{\color{red}{A-B}}.$$
For your case, we have
$$\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^{-1}=\dfrac{1}{\dfrac{4}{3}-\dfrac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Help in finding the integral function. Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
| Remember that $x^3+1 = (x+1)(x^2-x+1)$. That quadratic polynomial cannot be factored using real numbers, since the discriminant ($b^2-4ac\vphantom{\dfrac\int\int}$) is negative. So
$(x^3+1)^3 = (x+1)^3 (x^2-x+1)^3$. That should tell you what the partial fraction decomposition is.
To handle $x^2-x+1$, one (of course) completes the square:
\begin{align}
x^2-x+1 & = \left( x^2 - x + \frac 1 4\right) + \frac 3 4 \\[10pt]
& = \left( x - \frac 1 2 \right)^2 + \frac 3 4 \\[10pt]
& = \frac 3 4 \left( 1 + \left( \frac{2x-1}{\sqrt 3} \right)^2 \right).
\end{align}
Then $\tan\theta = \dfrac{2x-1}{\sqrt 3}$ and $1+\tan^2\theta=\sec^2\theta$ and $\sec^2\theta\,d\theta = \dfrac{2\,dx}{\sqrt 3}$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve $2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$ $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$
My try:
$$ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = \sqrt{4}\left(\frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} + \frac{1}{2}\cos{\frac{3x}{2}}\right) = 2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right)
$$
$$
2+ 2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) = 4\sin^2{\frac{x}{4}}
$$
$$
2\sin\left({\frac{3x}{2} + \frac{\pi}{6}}\right) + \cos{\frac{x}{2}}=0
$$
And then...
| We have $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$ $$1+\frac{1}{2}\cos{\frac{3x}{2}} + \frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} = \frac{4}{2}\sin^2{\frac{x}{4}}$$
$$\cos{\frac{3x}{2}}\cos \frac{\pi}{3} +\sin{\frac{3x}{2}}\sin\frac{\pi}{3} =-\left(1-2\sin^2{\frac{x}{4}}\right)$$
$$\cos\left(\frac{3x}{2}-\frac{\pi}{3}\right)=-\cos \frac{x}{2}=\cos \left(\pi-\frac{x}{2}\right)$$ Now, writing the general solution as follows $$\frac{3x}{2}-\frac{\pi}{3}=2n\pi\pm \left(\pi-\frac{x}{2}\right)$$ taking positive sign, we get $$\frac{3x}{2}-\frac{\pi}{3}=2n\pi+ \left(\pi-\frac{x}{2}\right)$$ $$2x=2n\pi+ \frac{4\pi}{3}$$ $$\color{blue}{ x=n\pi+\frac{2\pi}{3}}$$ taking negative sign, we get
$$\frac{3x}{2}-\frac{\pi}{3}=2n\pi- \left(\pi-\frac{x}{2}\right)$$
$$x=2n\pi-\pi+\frac{\pi}{3}$$
$$\color{blue}{ x=2n\pi-\frac{2\pi}{3}}$$ Where, $n$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate derivative of integral I tried to calculate the derivative of this integral:
$$\int_{2}^{3+\sqrt{r}} (3 + \sqrt{r}-c) \frac{1}{2}\,{\rm d}c $$
First I took the anti-derivative of the integral:
$$\frac{1}{2}\left(\frac{-c^2}{2}+c\sqrt{r}+3c\right)$$
Then I evaluated the integral:
$$-\frac{1}{2}\left(\frac{3+\sqrt{r})^2}{2} + (3 + \sqrt{r})\sqrt{r}+3(3+\sqrt{r})\right)-\frac{1}{2}\left(-\frac{(2^2)}{2}+2\sqrt{r}+3\cdot 2\right)$$
After I simplified I got:
$$\frac{1 + 2\sqrt{r}+r}{4}$$
I should get:
$$\frac{1 + \sqrt{r}}{4\sqrt{r}}$$
But I cannot get this result.
Can someone help? what am I missing?
| As it appears the derivative is taken with respect to $r$. The integral in question is given by, and evaluated as, the following:
\begin{align}
I(r) &= \frac{1}{2} \, \int_{2}^{3+\sqrt{r}} (3 + \sqrt{r} - t) \, dt \\
&= \frac{1}{2} \, \left[ (3 + \sqrt{r}) t - \frac{t^{2}}{2} \right]_{2}^{3 + \sqrt{r}} \\
I(r) &= \frac{(3 + \sqrt{r})^{2}}{4} - (2 + \sqrt{r}).
\end{align}
Now by differentiation
\begin{align}
\frac{dI}{dr} &= \frac{2}{4} \cdot (3 + \sqrt{r}) \cdot \frac{1}{2 \sqrt{r}} - \frac{1}{2 \sqrt{r}} \\
&= \frac{3 + \sqrt{r}}{4 \sqrt{r}} - \frac{1}{2 \sqrt{r}} = \frac{1 + \sqrt{r}}{4 \, \sqrt{r}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show divisibility by 7 I was stuck at this question:
Suppose $a^2+b^2=c^2$ for $a,b,c \in \mathbb Z$, and neither $a$ nor $b$ is a multiple of 7. Show that $a^2-b^2$ is a multiple of 7
I tried to write $b^2$ as $c^2-a^2$ then get $a^2-b^2=2a^2-c^2$. But this does not seem to generate the solution.
How to solve problems like this, am I missing some theorems concerning Pythagoras numbers?
| Using Euclid's formula, $a=2mn, b=m^2-n^2$
We have $7\nmid2mn(m^2-n^2)$
Now, $(m^2-n^2)^2-(2mn)^2=m^4+n^4-6m^2n^2\equiv m^4+n^4+m^2n^2\pmod7$
But $(m^2-n^2)(m^4+n^4+m^2n^2)=(m^2)^3-(n^2)^3\equiv1-1\pmod7$ using Fermat's Little Theorem as $(m,7)=(n,7)=1$
$\implies7|(m^4+n^4+m^2n^2)$ as $7\nmid(m^2-n^2)$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry identity $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$ How to prove that $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$?
I tried manupulating the left hand side but ended up in $\frac{\cos x}{\sin^2 x}$.
Can someone show me? Thanks in advance.
| $$\csc(x)\cot(x)=\frac{\cos^3(x)}{\sin^2(x)}+\cos(x)\Longleftrightarrow$$
$$\csc(x)\cot(x)=\frac{\cos^3(x)+\cos(x)\sin^2(x)}{\sin^2(x)}\Longleftrightarrow$$
$$\csc(x)\cot(x)\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\frac{1}{\sin(x)}\cdot\frac{\cos(x)}{\sin(x)}\cdot\sin^2(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\cos(x)=\cos^3(x)+\cos(x)\sin^2(x)\Longleftrightarrow$$
$$\cos(x)=\cos(x)(\cos^2(x)+\sin^2(x))\Longleftrightarrow$$
$$1=\cos^2(x)+\sin^2(x)\Longleftrightarrow$$
$$1=1$$
The left hand side and right hand side are identical -> identity has been verified
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x\rightarrow \infty $ I want to calculate limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x \rightarrow \infty $ without L'Hôpital's rule.
I found this task on the Internet. The answer given by the author is $2$.
$$\lim_{x\rightarrow\infty} x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2}) = \\
= \lim_{x\rightarrow\infty} x \cdot \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot(\sqrt{x^2+3}-\sqrt{x^2+2}) = (*)\\
$$
at this point I took care of a sine interior
$$ \lim_{x\rightarrow\infty} \sqrt{x^2+3}-\sqrt{x^2+2} = \\
=\lim_{x\rightarrow\infty} \frac{(\sqrt{x^2+3}-\sqrt{x^2+2}) \cdot(\sqrt{x^2+3} + \sqrt{x^2+2})}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\
= \lim_{x\rightarrow\infty} \frac{1}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \left[\frac{1}{\infty}\right] = 0 \\
$$
and whole sin function
$$ \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}} = \left[\frac{\sin(y\rightarrow0)}{y\rightarrow0}\right] = 1 $$
finally
$$(*) =\lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot \frac{x}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\ = \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot \frac{x}{x \left(\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{1}{x^2}}\right)} = \frac{1}{2} $$
Unfortunately $\frac{1}{2} \neq 2$ so who is wrong? (and if I'm wrong, then why?)
| It would be absurd not to agree with everything said, but this is how I would approach this task:
$\begin{aligned}\lim\limits_{x\to\infty}\left(x\cdot\sin\left(\sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim\limits_{x\to\infty}\frac{\sin\left(x\left(\sqrt{1+\frac3{x^2}}-\sqrt{1+\frac2{x^2}}\right)\right)}{\frac1x}\\&=\lim\limits_{x\to\infty}\frac{\sin\left(\frac{\frac1{x}}{\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}}\right)}{\frac1{x}}\\&=\lim\limits_{x\to\infty}\frac{\sin\left(\frac1{x\left(\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}\right)}\right)}{\frac1{x\left(\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}\right)}}\cdot\frac1{\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}}\\&=\frac12\end{aligned}$
Which is essentially the same thing you did.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to verify my solution to an separable differential equation? I have this question:
Find the general solution to the separable differential equation
$$
\frac{dy}{dx} = y(1-y).
$$
My attempt is :
$$
\frac{dy}{y(1-y)} = dx
$$
$$
\frac{1}{y(1-y)} = \frac{A}{y}+\frac{B}{(1-y)}
$$
$$
1=A(1-y)+ B y
$$
$$
A=1, B-A=0, so B=1
$$
$$
\frac{dy}{y} + \frac{dy}{1-y} = dx
$$
$$
ln(y)-ln(1-y) =x+k
$$
$$
\frac{y}{(1-y)}=K^x
$$
$$
\frac{(1-y)}{y}=c^{-x}
$$
$$
\frac{1}{y-1}=c^{-x}
$$
$$
y=\frac{1}{1+c^{-x}}
$$
Firstly is my attempt correct is or is there a better/simpler/easier way of solving this?
And secondly how would I formally verify my solution?
| $$
\frac{dy}{y} + \frac{dy}{1-y} = dx\Longrightarrow \frac{dy}{y} - \frac{dy}{y-1} = dx
$$
But now
$$
\ln|y| - \ln|y-1| = x + C\Longrightarrow \ln \left|\frac{y}{y-1}\right| = x + C\Longrightarrow \left|\frac{y}{y-1}\right| = e^{x + C} = C_1e^x
$$
$C_1$ must be positive, but
$$
\frac{y}{y-1} = \pm C_1e^x = De^x,
$$
and $D$ may be negative or positive constant. Now
$$
\frac{y-1}{y} = \frac{1}{De^x}=1 - \frac1y \Longrightarrow y = \frac{1}{1-\frac{1}{De^x}} = \frac{De^x}{De^x - 1}
$$
And don't forget that $y\equiv 0$ and $y\equiv 1$ are solutions too.
| {
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"url": "https://math.stackexchange.com/questions/1383964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the approximate square root of $(a^4-a^{-4})/(a^2-a^{-2})$ Find the approximate square root of
$$\dfrac{\left( \dfrac{12}{5}\right)^4 - \left( \dfrac{5}{12}\right)^4 }{\left( \dfrac{12}{5}\right)^2 - \left( \dfrac{5}{12}\right)^2}$$
I tried using the formula for $(a^4-b^4)$ and $(a^2-b^2)$. Then cancelled the common $(a-b)$, substituted the values and simplified. Didn't work.
Answer is 13
Can someone tell me how to solve it with steps of possible?
It will be a great help
| First, this is straight equal to $\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2}$ through a simple cancellation. Now notice that
$$\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2} =\left(\frac{12}{5}\right)^{2} + \left(\frac{12}{5}\right)^{-2} = \left(\frac{5}{12}\right)^{2} \left( \left(\frac{12}{5}\right)^{4} +1 \right)$$
Now take
$$\sqrt{\left(\frac{5}{12}\right)^{2} \left( \left(\frac{12}{5}\right)^{4} +1 \right)} = \frac{5}{12}\sqrt{\left(\frac{12}{5}\right)^{4} +1 } \approx \frac{5}{12} \left(\frac{12}{5}\right)^{2} = \frac{12}{5} = 2.4$$
It's no problem to drop the $1$, because $1$ is very small in relation to $\left(\frac{12}{5}\right)^{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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For what $(a,b) \in R^+$ does $\int^\infty_b (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}})dx$ converge? For what pairs $(a,b) \in R^+$ does this integral converge?
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}} \right)dx
$$
| The integral converges iff $a = b$.
Let's rewrite the expression using big-O notation. So
$$\sqrt{x+a}-\sqrt{x} = x^{1/2}(\sqrt{1+a/x}-1) = x^{1/2}(1+a/2x+O(x^{-2})),$$
Therefore
$$\sqrt{\sqrt{x+a}-\sqrt{x}} = x^{1/4}(a/4x+O(x^{-2}))$$
and similarly
$$\sqrt{\sqrt{x}-\sqrt{x-b}} = x^{1/4}(b/4x+O(x^{-2})).$$
Hence
$$
\int\limits^{\infty}_{b} \left (\sqrt{\sqrt{x+a}-\sqrt{x} \vphantom{\sqrt{x}-\sqrt{x-b}}}-\sqrt{\sqrt{x}-\sqrt{x-b}} \right) dx = \int\limits^{\infty}_{b} \left (x^{1/4}((a-b)/4x + O(x^{-2})\right)dx = \int\limits^{\infty}_{b} \left (x^{1/4}O(x^{-2})\right)dx + \frac{(a-b)}{4}\int\limits^{\infty}_{b} \left (x^{-3/4}\right)dx = \int\limits^{\infty}_{b} \left (O(x^{-7/4})\right)dx + \frac{(a-b)}{4}\int\limits^{\infty}_{b} \left (x^{-3/4}\right)dx.
$$
The first integral converges while the second integral doesn't unless $a=b$.
NOTE $O(x^n)$ means bounded by $ax^n$ ($a$ is a constant) and thus $(1+x)^{1/2}=1+x/2+O(x^2)$ for $|x| < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ (a+b) \cos(x) = a-b$, if $ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$ $ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$ Prove that $ (a+b) \cos(x) = a-b$, if $$ \sin(x) + \tan(x) = \frac{4a\sqrt{ab}}{a^2-b^2},$$
$$ \tan(x) - \sin(x) = \frac{4a\sqrt{ab}}{a^2-b^2}.$$
I tried solving it with system, but with no result.
| As written, the solution to the system of equations is $\sin x = 0$. I think you want
$$\tan x -\sin x={4b\sqrt{ab}\over a^2-b^2}.$$ Then solving the system of equations yields
$$\sin x =\frac12 \left({4a\sqrt{ab}\over a^2-b^2}-{4b\sqrt{ab}\over a^2-b^2}\right)={2\sqrt{ab}\over a+b},$$
which gives $$\cos^2 x = 1 -\sin^2 x = {(a-b)^2\over(a+b)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:$2 \cosh 2x - \sinh 2x =2$
Express $\cosh 2x$ and $\sinh 2x$ in exponential form and hence solve for real values of $x$ the equation:
$2 \cosh 2x - \sinh 2x =2$
Here is my idea:
$$2 \cosh 2x- \sinh 2x = \frac{2(e^x)^2+2(e^{-x})^2}{2} - \frac{(e^x)^2-(e^{-x})^2}{2}$$
$$2=\frac{(e^x)^2+3(e^{-x})^2}{2}$$
$$4 = (e^x)^2 + 3(e^{-x})^2$$
$$4= (e^x)^2 + \frac{(3)(1)}{(e^x)^2 }$$
Multiplying both sides by $(e^x)^2$
$$4(e^x)^2= (e^x)^4 +3$$
or
$$(e^x)^4 - 4(e^x)^2 = -3$$
This sort of looks like something I could solve by completing the square or some other technique for solving a quadratic. But this is where I am stuck.
I know that $x=0$ and $x=0.549$ are the solutions.
| Hint: i have $$e^{2x}+e^{-2x}-1/2e^{2x}+1/2e^{-2x}=2$$ multiplying by $e^{2x}$ we obtain $$e^{4x}+1-1/2e^{4x}+1/2=2e^{2x}$$ thus we get $$e^{4x}-4e^{2x}+3=0$$
with $$u=e^{2x}$$ we get $$u^2-4u+3=0$$
$u_1=3$ or $u_2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation Algebra: $\sum_{n=2}^\infty \frac{1}{2^{n-1}}$ What rule do we use to get the numerical value of $$\sum_{n=2}^\infty \frac{1}{2^{n-1}}$$
My notes give the answer straight away, without explaining what the steps are.
// Thanks everyone!
| $$\sum_{n=2}^\infty \frac{1}{2^{n-1}}=\frac{1}{2}\sum_{n=2}^\infty \frac{1}{2^{n-2}}=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+...\right)=\frac{1}{2}\frac{1}{1-\frac{1}{2}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the equation of the tangent to $ \ y= \frac{1}{x-4} $ at the point L where $\ x=b \ $ This is what I've done so far:
$$y' \ = \ -\frac{1}{(x-4)^ 2} \ \ \rightarrow \ \
y' (b) \ = \ -\frac{1}{( b-4)^2 } $$
Not sure what to do after this. Am I doing it right?
| Notice, we have $$y=\frac{1}{x-4}$$ substituting $x=b$, we get y-coordinate of the point L as follows $$y=\frac{1}{b-4}$$
The slope of the tangent at general point
$$\frac{dy}{dx}=\frac{-1}{(x-4)^2}$$ Hence, the slope at the point $L\left(b, \frac{1}{b-4} \right)$
$$\frac{dy}{dx}=\frac{-1}{(b-4)^2}$$
The equation of a atraight line passing through the point $(x_1, y_1)$ & having a slope $m$ is given as $$\color{blue}{y-y_1=m(x-x_1)}$$
Hence the equation of the tangent passing through the point $L\left(b, \frac{1}{b-4} \right)$ & having slope $\frac{-1}{(b-4)^2}$ $$y-\frac{1}{b-4}=\frac{-1}{(b-4)^2}(x-b)$$ $$(b-4)^2y-(b-4)=-x+b$$
$$x+(b-4)^2y+4-2b=0$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the line:}\ x+(b-4)^2y+4-2b=0}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Estimate for $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ If the triangle ABC has sides $a,b,c$ opposite to the vertices A,B,C respectively and $\Delta$ is the area.The expression $\frac{1}{\Delta}\sqrt{abc(a+b+c)}$ is
$(A)\leq16\hspace{1 cm}(B)\geq16\hspace{1 cm}(C)\leq4\hspace{1 cm}(D)\geq4$
Using Herons formula, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ and $s=\frac{a+b+c}{2}$, given expression reduces to
$4\sqrt{\frac{abc}{(a+b-c)(b+c-a)(a+c-b)}}$ and I could not solve further to find the answer. Please help....
| with $a=y+z,b=x+z,c=x+y$ we get
$$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}$$
by AM-GM we have $$(x+y)(x+z)(y+z)\geq 8xyz$$
thus our term above is $$\sqrt{2} \sqrt{\frac{(x+y) (x+z) (y+z)}{x y z}}\geq \sqrt{8}\sqrt{2}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad \quad \quad \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$
By observation if all the three value
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.
| Plugging in $a = b = c$, we get
\begin{align*}
3 = \frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}= 3\frac{m - a^2}{2a^2}
&\implies m - a^2 = 2a^2 \\
&\implies m = 3a^2.
\end{align*}
On the other hand, plugging $a = b = c$ into the answers we get
\begin{align*}
&\boldsymbol{(1)} \quad - a^2 \\
&\boldsymbol{(2)} \quad a^2 \\
&\boldsymbol{(3)} \quad 2 a^2 \\
&\boldsymbol{(4)} \quad 3 a^2 \\
\end{align*}
so the answer must be $\boxed{\boldsymbol{(4)}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarly solving the above equation it comes out to be:
$$\sqrt2 u^2-u-\sqrt2=0=(u-\sqrt2)(\sqrt2u+1)=0 \implies \sin x+\cos x=\sqrt2 \quad(1)$$
and
$$\sin x+\cos x= \frac{-1} {\sqrt2}\quad (2)$$
so solving the results differently i got the answers:
$$x=2n\pi + \frac{\pi}4, 2n\pi +\frac{7\pi}{12}, 2n\pi-\frac{\pi}{12}$$
but the answers are:
$$x=2n\pi + \frac{\pi}4, 2n\pi -\frac{5\pi}{12}, 2n\pi+\frac{11\pi}{12}$$
I divided the eq(2) by $\sqrt2$
| The equation can be rewritten
$$\sqrt2\sin(x-\frac\pi4)=\sqrt2\sin(2x),$$
hence
$$x-\frac\pi4=2x+2k\pi\text{ or }x-\frac\pi4=\pi-2x+2k\pi,$$
$$x=\frac{24k-3}{12}\pi\text{ or }x=\frac{8k+5}{12}\pi.$$
The second formula covers the first, so that
$$\color{green}{x=\frac{8k+5}{12}\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2 \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after that i can't get the expansion to work.
Can you help me out?
| Starting from what you ought to have:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (2+\frac{3}{2}i)^2$.
we get:
$(z−(2+\frac{3}{2}i))^2 = -7-i + (4+6i-\frac{9}{4}) = \frac{-21+20i}{4} = (\frac{2+5i}{2})^2$.
The last step I obtained by guessing. If you want a systematic way to find the square root of a complex number if it is a perfect square (the square of a rational complex number) you can use the following technique.
Given $(a+bi)^2 = c+di$ where $a,b,c,d \in \mathbb{R}$:
$a^2-b^2 = c$ and $2ab = d$ [by comparing real and imaginary parts].
Thus $4 a^2 - 4 a^2 b^2 = 4 a^2 c^2$ and $4 a^2 b^2 = d^2$ [multiply the first by $2a^2$ and square the second].
Thus $4 a^4 - 4c a^2 - d^2 = 0$ [Add them together to get a quadratic in $a$].
Thus $(2a^2-c)^2 = c^2+d^2$ [Complete the square].
Thus $2a^2-c = \sqrt{c^2+d^2}$ and hence $a = \pm \sqrt{\frac{\sqrt{c^2+d^2}+c}{2}}$.
Also $2b^2 = 2a^2 - 2c = \sqrt{c^2+d^2}-c$ and hence $b = \pm \sqrt{\frac{\sqrt{c^2+d^2}-c}{2}}$.
[Note that the choices of sign are dependent, so it may be better to use $b = \frac{d}{2a}$ instead.]
In the above case we get:
$\sqrt{-21+20i} = \pm \left( \sqrt{\frac{\sqrt{21^2+20^2}-21}{2}} + \sqrt{\frac{\sqrt{21^2+20^2}+21}{2}} i \right) = \pm (2+5i)$.
[Here I have chosen the signs correctly.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How compute $\cos(5\theta)$ and $\sin(5\theta)$? I would like to compute $\cos(5\theta)$ and $\sin(5\theta)$. I can use the formula $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ and $\sin(a+b)=\sin(a)\cos(b)+cos(a)\sin(b)$ but it's a little bit to long. Is there an other way to compute it ?
| \begin{align}
c_2 &=c^2-s^2, \hspace{5mm} s_2=2cs \\
c_4 &= c_2^2-s_2^2 = (c^2-s^2)^2-4c^2s^2 = c^4-6c^2s^2+s^4,\\
s_4 &= 2c_2s_2 = 4cs(c^2-s^2) = 4c^3s-4cs^3. \\
c_5 &= c_4c-s_4s = (c^4-6c^2s^2+s^4)c-(4c^3s-4cs^3)s = c^5-10c^3s^2+5cs^4,\\ s_5 &= c_4s+s_4c = (c^4-6c^2s^2+s^4)s+(4c^3s-4cs^3)c = 5c^4s-10c^2s^3+s^5.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of Legendre function I'm currently trying to solve the sum
$$
f(x)=\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}P_n(x),
$$
where $P_n(x)$ is the Legendre function of order n.
I also named the sum $f(x)$ since I'm the solution will be a function of $x$. The first thing that can be noticed is that the function $f(x)$ is an uneven function.
My first step was to take the derivative of $f(x)$, this then yields that:
$$
f'(x)=\sum\limits_{n=0}^\infty x^{n}P_n(x)+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x).
$$
Using the generating function
$$
\frac{1}{\sqrt{1-2hz+h^2}}=\sum\limits_{n=0}^\infty h^nP_n(z)
$$
I'm able to rewrite the first term as:
$$
f'(x)=\frac{1}{\sqrt{1-x^2}}+\sum\limits_{n=1}^\infty\frac{x^{n+1}}{n+1}P'_n(x).
$$
Now for the second term I'm still stuck, I've been trying a few recursive relations, in the hope to maybe get a first order differential equation for $f(x)$. So far I haven't found the right recursive relation yet.
Now I'm wondering if there is maybe a better approach that I should have taken, or that if there is an easy trick to solve the second term?
After playing with the function in Mathematica I figured out that the most probable solution is
$$
f(x)=\mathrm{arctanh}(x),
$$
this gives me a direction, but still I have no clue for the second term. All help is welcome!
| Let:
$$f(x) = \sum _{n=0}^{\infty} \frac{x^{n+1} \, P_{n}(x)}{n+1}$$
As you said:
$$f^{'}(x) = \sum_{n=0}^{\infty} x^{n} \, P_{n}(x) + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \, P_{n}^{'}(x)$$
And using the generating function you showed, we get:
$$f^{'}(x) = \frac{1}{\sqrt{1-x^{2}}} + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } \, P_{n}^{'}(x)$$
Now, note that:
$$
{ P }_{ n }(x)'\quad =\quad \frac { \left( n+1 \right) \left( x{ P }_{ n }(x)\quad -\quad { P }_{ n+1 }(x) \right) }{ 1-{ x }^{ 2 } }
$$
So we get:
\begin{align}
f^{'}(x) &=\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } +\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n+1 } }{ n+1 } } \frac { \left( n+1 \right) \left( x{ P }_{ n }(x) - { P }_{ n+1 }(x) \right) }{ 1-{ x }^{ 2 } } \\
&= \frac {1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty}{x^{n+1} \left(x {P}_{n}(x) - {P}_{n+1}(x) \right)} \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+2} \left( P_{n}(x) \right) - \frac{1}{1-x^{2}} \sum_{n=0}^{\infty} x^{n+1} \, P_{n+1}(x) \\
&= \frac{1}{\sqrt{1-x^{2}}} +\frac{x^{2}}{1-x^{2}} \sum_{n=0}^{\infty} x^{n} \left(P_{n}(x) \right) - \frac{1}{1-x^{2}}\left(\sum_{n=0}^{\infty} x^{n} \, P_{n}(x) -1 \right) \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}}{1-x^{2}} \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{1-x^{2}} \left(\frac{1}{\sqrt{1-x^{2}}} -1\right) \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{x^{2}-1}{ (1-x^{2}) \, \sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} \\
&= \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} + \frac{1}{1-x^{2}} = \frac {1}{1-x^{2}}
\end{align}
Integration yields:
\begin{align}
f(x) &= \int \frac{dx}{1-x^{2}} = \frac{1}{2} \, \int \left(\frac{1}{1-x} + \frac{1}{1+x} \right) dx \\
&= \frac{1}{2} \, \left[\ln(1+x)-\ln(1-x)\right] + C \\
&= \frac{\ln\left(\frac{1+x}{1-x}\right)}{2} + C
\end{align}
and for $x=0$ we have $f(0) = 0$ so we get $C=0$.
Hence our answer is:
$$f(x) = \sum_{n=0}^{\infty}{\frac{{x}^{n+1}}{n+1} \, P_{n}(x)} = \frac{1}{2} \, \ln\left(\frac{1+x}{1-x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| Try this:
$$3 \cos x + 2\sin x=\sqrt{13}\left(\frac{3}{\sqrt{13}}\cos x+\frac{2}{\sqrt{13}}\sin x\right)\\
=\sqrt{13}\sin(\arcsin\frac{3}{\sqrt{13}}+x)=1$$
You can try to solve it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Simplifying integral:$\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u$ $$I(t)=\int_{0}^{\infty}{\exp\left(-\left(u^2+{ {\alpha^2}\over {16u^2t}}\right)\right)}~\mathrm{d}u $$
where $\alpha$ and $t$ are positive constant.
P.S.I would like to edit this problem, because I found my mistake about inner sign.
Thanks!
| Using
$$x^{2} - \frac{a^{2}}{x^{2}} = \left(x - \frac{i a}{x} \right)^{2} + 2 i a $$
then
\begin{align}
I &= \int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du
= e^{-2 i a} \, \int_{0}^{\infty} e^{- \left(u - \frac{i a}{u}\right)^{2}} \, du
\end{align}
where $4 \sqrt{t} \, a = \alpha$. Let $u = \sqrt{i a} x$ to obtain
\begin{align} \tag{1}
e^{2 i a} \, I = \sqrt{i a} \, \int_{0}^{\infty} e^{- i a \, \left(x - \frac{1}{x}\right)^{2}} \, dx.
\end{align}
Now let $x \to \frac{1}{x}$ to obtain
\begin{align}\tag{2}
e^{2 i a} \, I &= \sqrt{i a} \, \int_{0}^{\infty} e^{-i a \, \left( x - \frac{1}{x} \right)^{2}} \, \frac{dx}{x^{2}}.
\end{align}
Adding equations (1) and (2) lead to
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, \left(x - \frac{1}{x}\right)^{2}} \, \left(1 + \frac{1}{x^{2}}\right) \, dx.
\end{align}
Let $y = x + \frac{1}{x}$ to obtain
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \int_{0}^{\infty} e^{-i a \, y^{2}} \, dy
\end{align}
Set $y = \sqrt{\frac{x}{i a}}$ to obtain
\begin{align}
\frac{2 \, e^{2 i a}}{\sqrt{i a}} \, I = \frac{1}{2} \, \sqrt{\frac{\pi}{i a}}
\end{align}
From this the integral is given by
\begin{align}
\int_{0}^{\infty} e^{- \left(u^{2} - \frac{a^{2}}{u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{-2 i a}.
\end{align}
When $4 \sqrt{t} a = \alpha$ then
\begin{align}
\int_{0}^{\infty} e^{- \left(u^{2} - \frac{\alpha^{2}}{16 \, t \, u^{2}}\right)} \, du = \frac{\sqrt{\pi}}{4} \, e^{- \frac{i \, \alpha}{2 \sqrt{t}}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$
and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n$ becomes sufficiently large. I am wondering whether showing the ratio between the two terms approaching $1$ is sufficient to show the difference is $0$; if so, how can this be formalized? If not, how can it more directly be shown that the difference approaches $0$?
| Hint: If we can set $a=(n+1)^{\frac 17}$ and $b=n^{\frac 17}$, then the expression becomes $$\lim_{n \to \infty} [a-b]=0.$$ It suffices to prove that $$a-b \to 0$$ as $n \to \infty$.
The Binomial Theorem at degree $7$ gives us
$$a^7-b^7=(a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6). \tag{$\star$}$$
We have
\begin{align}
a-b
&= \frac{a^7-b^7}{a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6} \quad \text{by } (\star)\\
&= \frac 1{a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6} \\
&\to 0
\end{align}
as $n \to \infty$.
Explanations for the above steps:
*
*$a^7-b^7=[(n+1)^{\frac 17}]^7-[n^{\frac 17}]^7=(n+1)-n=1$
*$a =(n+1)^{\frac 17} \to \infty$ and $b = n^{\frac 17} \to \infty$
*The variables $a$ and $b$ are in only the denominator for the penultimate step above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the system of equations $\begin{cases} xy-2y-3 &=\sqrt{y-x-1}+\sqrt{y-3x+5} \\ (1-y)\sqrt{2x-y}+2(x-1) &=(2x-y-1)\sqrt{y}. \end{cases}$ Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases}
xy-2y-3 &=\sqrt{y-x-1}+\sqrt{y-3x+5} \\
(1-y)\sqrt{2x-y}+2(x-1) &=(2x-y-1)\sqrt{y}.
\end{cases}$$
I think this system of equations have no solution, but I can't prove it.
I really appreciate if some one can help me. Thanks!
| Let me try. One has $y \geq 0$ and $y \leq 2x$.
Then, the second equation can be written as follows:
$$(1-y)\sqrt{2x-y} + (2x-1) -1 - (2x-1)\sqrt{y} + y\sqrt{y} = 0$$
$$(1-\sqrt{y})\left( (1+\sqrt{y})\sqrt{2x-y} + (2x-1) - 1 - \sqrt{y} - y\right) = 0$$
$$(1-\sqrt{y})(\sqrt{y}\sqrt{2x-y}-\sqrt{y} + \sqrt{2x-y} - 1 + (2x-y) -1) = 0$$
$$(1-\sqrt{y})(\sqrt{2x-y}-1)(\sqrt{y}+1+\sqrt{2x-y}+1) = 0$$
Thus, one has $\sqrt{y} = 1$ or $\sqrt{2x-y} = 1$.
The first case: $y = 1$. From this and the first equation, one has
$$x-5 = \sqrt{-x} + \sqrt{6-3x}.$$ But, $x\geq \frac{y}{2} \geq 0$. Then, there is no solution for this case.
The second case: $2x-y = 1$ or $y = 2x-1$. From this and the first equation, one gets:
$$2x^2 - 5x -1 = \sqrt{x-2} + \sqrt{4-x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Solve the equation: $(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$ Solve the equation:
$(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$
I used wolframalpha.com and got only solution $x=-\dfrac{1}{3}$.
And this is my try:
Condition: $x\ge-\dfrac{2}{3}$.
The equation is equivalent to:
$\quad(3x+1)^2\sqrt{3x+2}+9-9\sqrt{3x+2}=27x^2-6x-5+3\sqrt{10+3x}-9$
$\Leftrightarrow (3x+1)^2\sqrt{3x+2}-\dfrac{9(3x+1)}{1+\sqrt{3x+2}}=(3x+1)(9x-5)+\dfrac{3(3x+1)}{\sqrt{10+3x}+3}$
$\Leftrightarrow (3x+1)\left[(3x+1)\sqrt{3x+2}-\dfrac{9}{1+\sqrt{3x+2}}-9x+5-\dfrac{3}{\sqrt{10+3x}+3}\right]=0$
But I can't prove that $(3x+1)\sqrt{3x+2}-\dfrac{9}{1+\sqrt{3x+2}}-9x+5-\dfrac{3}{\sqrt{10+3x}+3}>0$.
So, who can help me solve it?
| I think the polynomial that you get by multiplying by the three conjugates is
$$59049x^{10}-118098x^9+164025x^8-236196x^7 -441774x^6+133164x^5+373086x^4-390636x^3 -193203x^2+141054x+50641=0$$
This has four real zeros, including $x=-1/3$, but those would include negative signs for the square-roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand it now.
| Observe $$\left(x-\frac{1}{3}\right)^2=x^2-\frac{2}{3}x+\frac{1}{9}$$$$=\frac{1}{3}\left(3x^2-2x\right)+\frac{1}{9}.$$
This is almost the original expression, we're just missing a $7$. Then,
$$\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}.$$
Now, use the original equality to simplify.
Then, we get
$$
\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}$$ $$=-\frac{7}{3}+\frac{1}{9}
$$ $$=-\frac{20}{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 4
} |
Show that $\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$.
My problem is to show that
$$\frac{x}{3!}-\frac{x^3}{5!}+\frac{x^5}{7!}-\cdots\leq \frac{1}{\pi}$$
for all $x\in\Bbb R$.
I was thinking of first finding the max and then show that its less than $1/\pi$. But it is hard to find it. I get that the series is equal to
$$f(x)=\frac{x-\sin x}{x^2}.$$
Then,
$$f'(x)=\frac{1-\cos x}{x^2}-\frac{2(x-\sin x)}{x^3}=0$$
if and only if
$$2\sin x=x(1+\cos x),$$
which I am unable to solve, appart from the obvious solutions $x=0$ and $x=\pi$. But if $x=\pi$ is the max, then we are done because $f(\pi)=1/\pi$.
| Note that $f$ is odd. Thus, it suffices to look at the interval $[0, \infty)$.
Lemma: For $x$ > $\pi$, we have $$\sin(x) > \frac{(\pi^2-x^2)x}{\pi^2+x^2}.$$
Proof: See
here.
Thus, for $x > \pi$, we have
$$
\frac{1}x - \frac{\sin(x)}{x^2} < \frac{1}x + \frac{\pi^2-x^2}{x(\pi^2+x^2)} := g(x).$$
For $x > \pi$ (or even $x > 0$), the RHS is monotonically decreasing. Thus, we can conclude that for $x \in (\pi, \infty),$ we have $$f(x) < g(\pi) = \frac{1}{\pi}.$$
Now we will find the maximum of $f$ on $[0, \pi]$. First, we check the endpoints, $$\lim_{x \rightarrow 0} f(x) = 0$$ and $f(\pi) = \frac{1}{\pi}$.
Now we find all the extremas on our desired interval. Taking the derivative and setting it equal to $0$ gives $$2\sin(x)-x \cos(x) - x = -2 \cos(x/2)(x \cos(x/2)-2\sin(x/2)) = 0.$$ The only root of the first factor in our interval is at $x = \pi$. Now we consider $$x \cos(x/2)-2\sin(x/2) = \sqrt{x^2+4}\sin(x/2-\tan^{-1}(2/x)) = 0.$$ The roots of $\sin(\theta)$ are at multiples of $\pi$. Note that $$(x/2-\tan^{-1}(x/2))' = \frac{x^2+8}{2x^2+8}$$ so this function is strictly increasing for $x \ge 0$. Furthermore, $$\lim_{x \rightarrow 0^{+}} \frac{x}2- \tan^{-1}(2/x) = - \frac{\pi}2$$ and $$\frac{\pi}2 - \tan^{-1}(\pi/2) \approx 1$$ so $$\sin(x/2-\tan^{-1}(2/x))$$ has no zeroes in this interval. Thus, the global maximum is at $x = \pi$ which givse $f(\pi) = \frac{1}{\pi}$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
} |
Convergent/divergent series Is the following series divergent/convergent?
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks
| We should be careful here. It's tempting to group the terms into natural subsums and argue the grouped series converges and leave it at that. But the convergence of a grouped series converges does not imply the original series converges in general. The obvious example being $1 + (-1) + 1 + (-1) +\cdots.$ Here's an example with the terms $\to 0$:
$$1-1 +1/2+1/2 -1/2 - 1/2 + 1/3+1/3 +1/3 - 1/3 -1/3-1/3 + \cdots.$$
One sufficient condition for the convergence of the grouped series to imply convergence of the orignal series is the following: Suppose the original series is $\sum_n a_n$ and we know the grouped series
$$(a_1 + \cdots + a_{n_1}) + (a_{n_1+1}+ \cdots +a_{n_2 }) + \cdots $$
converges. If
$$\lim_{k\to\infty} (|a_{n_k+1}|+ \cdots +|a_{n_{k+1}}|) = 0,$$
then the original series converges. I'll leave the proof to the reader as it's not too bad. (Briefly, the condition shows the partial sums of the original series differ from those of the grouped series by an amount $\to 0,$ giving the desired convergence.)
For the problem at hand, this is fairly easy to check.
Here's my own proof for this specific problem, where I don't appeal to the $\log $ function at all. For the $k$th grouping of positive and negative terms, let $M=(2k-2)(2k-1)/2, N = (2k-1)(2k)/2.$ Then the grouping is
$$\tag 1 \frac{1}{M+1} + \cdots + \frac{1}{M+2k-1} - \frac{1}{N+1}-\cdots - \frac{1}{N+2k}$$ $$ = \sum_{j=1}^{2k-1} \frac{N-M}{(M+j)(N+j)} - \frac{1}{N+2k}.$$
Note that $N-M = 2k-1.$ In absolute value then, making absolutely brazen estimates, $(1)$ is $\le (2k-1)^2/(M+1)^2 + 1/N.$ This is on the order of $1/k^2$ as $k\to \infty.$ This implies the grouped series converges absolutely, hence converges. How about the original series? It converges by the first part of this answer, because if we slap absolute values on each term in $(1),$ the sum is is no more than $(4k-1)/M+1 \to 0$ as $k\to 0.$ This finishes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find $\lim\limits_{x\to\infty}\frac{(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)}{\left(1+x^{\frac{n+1}{2}}\right)^n}$ I have written a sequence $$(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)$$ as $$\frac{(x^{n+1}-1)(x^n+1)}{x-1}$$
Now, the limit is $$\lim\limits_{x\to\infty}\frac{(x^{n+1}-1)(x^n+1)}{(x-1)\left(1+x^{\frac{n+1}{2}}\right)^n}$$
What to do next?
| Dividing both numerator and denominator by $x\cdot x^2\cdot \ldots\cdot x^n=x^{\frac{n(n+1)}{2}}$ we have
\begin{align*}
\lim_{x\to\infty}\frac{(1+x)(1+x^2)(1+x^3)...(1+x^n)}{\left(1+x^{
\frac{n+1}{2}}\right)^n}&=\lim_{x\to \infty}\frac{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right)\ldots\left(1+\frac{1}{x^n}\right)}{\left(\frac{1}{x^{\frac{n+1}{2}}}+1\right)^n}\\
&=\frac{(1+0)(1+0)\ldots(1+0)}{(0+1)^n}\\
&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\to\infty}\frac{a^n}{n!}$ First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a riemann integral. Now I am thinking about sandwich theorem.
$$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$
Since $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$
$$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$
since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$
Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$
And by using sandwich theorem, $y=0$. Is this correct?
| Let $x_n=\frac{a^n}{n!}$.
$$\left|\frac{x_{n+1}}{x_n}\right|=\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=\frac{a^{n+1}n!}{a^n (n+1)!}= \frac{a}{n+1}\underset{n\to \infty }{\longrightarrow }0$$
and thus, by $x_n\to 0$ by Ratio test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the coefficients of a quadratic function with real roots cannot be in geometric progression
Suppose $$ax^2+bx+c$$ is a quadratic polynomial (where $a$, $b$ and $c$ are not equal to zero) that has real roots. Prove that $a$, $b$, and $c$ cannot be consecutive terms in a geometric sequence.
I tried writing the geometric sequence as $$a,\ b=ar,\ c=ar^2$$ and then substituting it back into the quadratic as $$ax^2+arx+ar^2$$ and then factoring and trying to prove that the discriminant was less than zero. But I ended up with $$r^2(x-2)(x+2)$$ which is not always less than zero. Any help would be appreciated.
| Note:
I added a proof that for odd $n$
the only root is $-1$.
Generalizing
Milo Brandt's answer,
which I thought of before I saw his,
this applies to a polynomial
of any even degree.
If the polynomial is of degree $2n$,
using his argument,
we need to find out
how many real roots
$p(x)
=x^{2n}+x^{2n-1}+...+x+1
$
can have.
But
$p(x)
=\frac{x^{2n+1}-1}{x-1}
$
has no real roots
because
the numerator and denominator
have the same sign
and at 1,
their common root,
$p(x) = 2n+1$.
For odd $n$,
the only real root is $-1$.
$n=3$ shows what happens;
I will then give the proof
for general odd $n$.
$x^3+x^2+x+1
=\frac{x^4-1}{x-1}
=\frac{(x^2+1)(x^2-1)}{x-1}
=\frac{(x^2+1)(x+1)(x-1)}{x-1}
=(x^2+1)(x+1)
$
for $x \ne 1$.
The only real root
is, obviously,
$x=-1$.
For general odd $n$,
since $n+1$ is even,
let $n+1 = 2^km$
where $m$ is odd.
Then,
just for $n=3$,
above,
$\begin{array}\\
x^n+x^{n-1}+...+x+1
&=\frac{x^{n+1}-1}{x-1}\\
&=\frac{x^{2^km}-1}{x-1}\\
&=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-1}m}-1)}{x-1}\\
&=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)(x^{2^{k-2}m}-1)}{x-1}\\
&=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)...(x^{2m}+1)(x^m+1)(x^m-1)}{x-1}\\
&=(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)...(x^{2m}+1)(x^m+1)\frac{x^m-1}{x-1}\\
\end{array}
$
Since $m$ is odd,
as proved above,
$\frac{x^m-1}{x-1}$
has no real roots.
All the terms
$x^{2^jm}+1$
for $j \ge 1$
are at least $1$
since the exponent is even.
Finally, since $m$ is odd,
$x^m+1$
has as its only real root
$x=-1$.
Therefore,
the whole polynomial
has $-1$ as its only real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
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Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
| This is equal to:
$$\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
which you can verify by multiplying your polynomial by $a+b+c$. So let's try to factor $a^3+b^3+c^3-3abc$.
So we know that
$$a+b+c\mid a^3+b^3+c^3-3abc$$
(The vertical line means "is a factor of.")
Now, let $\Omega$ be a cube root of unity (usually denoted by $\omega$ but i'll follow the notation in your question). Notice that the maps:
\begin{align}
a&\mapsto a\\
b&\mapsto b\Omega\\
c&\mapsto c\Omega^2\\
\end{align}
don't change $a^3+b^3+c^3-3abc$. Thus, applying them to the previous equation, we get:
\begin{align}
a+b+c&\mid a^3+b^3+c^3-3abc\\
a+b\Omega+c\Omega^2&\mid a^3+b^3+c^3-3abc\\
a+b\Omega^2+c\Omega&\mid a^3+b^3+c^3-3abc
\end{align}
These are its only factors, since it's a monic degree-3 polynomial and we've just supplied three monic degree-1 factors. Thus:
$$(a+b+c)(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)\\=a^3+b^3+c^3-3abc$$
Dividing out by $a+b+c$ and using the observation at the start of this answer, we get your factorization.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}
$$
My Attempt:
For $(\mathrm{i})$, We can write $\binom{n}{m}$ as the coefficient of $x^m$ in $(1+x)^n$. Thus we can also write $\binom{n-1}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-1}$ and $\binom{n-2}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-2}$. So we have to find the coefficient of $x^m$ in
$$
(1+x)^n+(1+x)^{n-1}+(1+x)^{n-2}+........+(1+x)^{m}
$$
Using the formula for the sum of a geometric progression, this sum equals
$$\frac{(1+x)^{n+1}-(1+x)^m}{x}$$
So we now need to find the coefficient of $x^m$ in
$$
\frac{(1+x)^{n+1}-(1+x)^m}{x}
$$
or, equivalently, we need to find the coefficient of $x^{m+1}$ in
$$
(1+x)^{n+1}-(1+x)^m = \binom{n+1}{m+1}
$$
We can use a similar method to solve $(\mathrm{ii})$. Can these questions be solved using combinatorial methods instead?
| First problem: We have $n+1$ different doughnuts (labelled $1$ to $n+1$) lined up in a row, and want to choose $m+1$ of them for breakfast. This can be done in $\binom{n+1}{m+1}$ ways. Let us count another way.
Maybe the leftmost doughnut chosen is $1$. There are $\binom{n}{m}$ ways to choose the rest.
Maybe the leftmost doughnut chosen is $2$. There are then $\binom{n-1}{m}$ ways to choose the others.
And so on.
Second Problem: This is done similarly. This time we are choosing $m+2$ doughnuts from $n+2$ doughnuts, labelled $1$ to $n+2$ and lined up in that order.
Look at the leftmost two doughnuts chosen. If the second one is Doughnut $2$, there are $\binom{n}{m}$ ways to choose the rest.
If the second one chosen is Doughnut $3$, there are $2$ ways to choose the first one, and $\binom{n-1}{m}$ ways to choose the rest, for a total of $2\binom{n-1}{2}$.
If the second one chosen is Doughnut $4$, there are $3$ ways to choose the first one, and $\binom{n-2}{m}$ ways to choose the rest, for a total of $3\binom{n-2}{m}$.
And so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $f’(x) = \sin x + (\sin4x)(\cos x)$, then $f’(2x^2 + \pi/2) $is? If $$f'(x) = \sin x + \sin4x \cdot \cos x,$$ then $$f'(2x^2 + \pi/2)$$ is?
Given answer: $$4x\cos(2x^2) – 4x\sin(8x^2) \sin(2x^2)$$
I tried and I'm getting the answer as $\cos(2x^2) - \sin(8x^2)\sin(2x^2)$
| $$f\'(x) = \sin x + \sin 4x \cos x$$
We need to find $f\'(2x^2+\pi/2)$.
Now
$$f\'(2x^2+π/2) = f\'(2x^2+π/2).Derivative of (2x^2+π/2)$$
$$ = {\sin (2x^2+π/2) + \sin 4(2x^2+π/2).Cos(2x^2+π/2)}.4x$$
$$ = 4x{\cos(2x^2) - \sin(8x^2).\sin(2x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
For a natural number $b$, $N(b)=$ number of natural numbers $ \ a \ $ such that the equation $x^2+ax+b=0$ has integral roots. For a natural number $b$, $N(b)= $ number of natural numbers $a$ such that the equation $x^2+ax+b=0$ has integral roots. What is the lowest possible value of $N(6)$?
| Rote approach. The solutions to $x^2+ax+6=0$ are $x_1=(1/2)(\sqrt{a^2-24}-a), x_2=(1/2)(-\sqrt{a^2-24}-a)$. First, $\sqrt{a^2-24}$ is rational iff it is integral, so we must have $a=\sqrt{c^2+24}$ for some integral ${c}$. Notice that if $c>12$ this expression can't be an integer, as the difference between consecutive square numbers is larger than 24 at that point, so you need only check $c = -4, -3, ..., 12$.
For a shortcut, note that if $x^2+ax+6=0$ then $x|6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| $$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum lies in the interval $(a_1,a_2)\cup(a_3,a_4)$
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
positive point of maximum lies in the interval
$(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$
I differentiated the equation $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ and put it equal to zero, to get $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$. Now what to do to get the desired interval. What is the significance of positive point of maximum in this question.
Please help. Thanks in advance.
| The zeroes of $f'$ are $-a+7\pm\sqrt{-14a+58}$. Now determine $a$ so that the zeroes are zero, you'll find $a=\pm3$. From here you'll be able to find the desired intervals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4cos^2(x)-2=\sqrt3$$
$$2(2\cos^2(x)-1)=\sqrt3$$
$$2\cos(2x)=\sqrt3$$
$$\cos(2x)=\frac{\sqrt3}{2}$$
$$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$
$$2x=\pm \frac{\pi}{6}+2n\pi$$
$$x=\pm \frac{\pi}{12}+n\pi$$
Considering the condition $\sin(x)\le 0$, we are left with
$$x=- \frac{\pi}{12}+n\pi$$
But the textbook's answer is
$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$
What did I overlook?
P.S. The problem and the answer from the texbook:
| $$2\sin^2x=2\cos^2x-\sqrt3$$
$$\iff2\sin^2x=2(1-\sin^2x)-\sqrt3$$
$$\iff\sin^2x=\dfrac{2-\sqrt3}4=\dfrac{(\sqrt3-1)^2}{8}$$
As $\sin x<0,\sin x=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left(\dfrac\pi6-\dfrac\pi4\right)$
$x=n\pi+(-1)^n\left(\dfrac\pi6-\dfrac\pi4\right)$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by induction that $\big(1+\frac{1}{n}\big)^n\le3-\frac{1}{n}$, as in
$\hspace{.4 in}$ A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
| Let $a_n = (1 + 1/n)^n.$
We want to show $a_{n+1} - a_{n} \geq \dfrac{1}{n(n+1)}$ for large $n$.
$\dfrac{a_{n+1}}{a_n} = \left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1}.$
The RHS can be expanded as
$\left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1} = \dfrac{n+1}{n} \times \left( \underbrace{1 - \dfrac{1}{n+1}} + \dfrac{1}{2!(n+1)^2}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} - \dfrac{1}{3!(n+1)^3}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) + \dots (-1)^{n+1} \dfrac{1}{(n+1)!(n+1)^{n+1}}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) \dots\left(1 - \dfrac{n}{n+1}\right)\right).$
Since $ \dfrac{n+1}{n} \times (1-\dfrac{1}{n+1}) = 1$, we have
$\dfrac{a_{n+1}}{a_n} = 1 + \dfrac{1}{2!(n+1)^2} - \dfrac{1}{3!(n+1)^3} \left(1 - \dfrac{2}{n+1} \right) + \dots$
So,
$|\dfrac{a_{n+1}}{a_n} - 1 - \dfrac{1}{2(n+1)^2}| \leq \dfrac{1}{3!(n+1)^3} + \dfrac{1}{4!(n+1)^4} + \dots \leq \dfrac{1}{6(n+1)^2n}.$
This implies $$ (n+1)^2 \left( \dfrac{a_{n+1}}{a_n} - 1 \right) \to 1/2$$ so upon multiplying the above by $na_n/(n+1)$ $$ n(n+1)(a_{n+1} - a_n) \to e/2 > 1.$$
Hence, $a_{n+1} - a_n \geq 1/n(n+1)$ for all large n.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
} |
$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?
| First recall some elementary algebra that if $A\cdot B = 0$, then $A = 0$ or $B = 0$ or both of them $A = 0, B = 0$. So $x(x^2-2) = 0 \to x = 0$ or $x^2-2 = 0$, and $x^2-2 = 0 \to (x-\sqrt{2})(x+\sqrt{2}) = 0 \to x-\sqrt{2} = 0$ or $x +\sqrt{2} = 0 \to x = \sqrt{2}$ or $x = -\sqrt{2}$. Thus there are $3$ solutions: $x = 0,-\sqrt{2}, \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Finding a function satisfying a certain inequality This is a continuation of this post where I tried to find a function $f(n)$ that would satisfy the induction step of an inductive argument and it was shown that such function does not exist.
Trying to fix the problem I've come up with a stronger inductive argument that now requires finding a more elaborate function.
More precisely I would like to find a function $f(n,k)$ such that the following relation would hold for any $0 \leq p \leq (n-2-l)/2$ and $l \leq k$, $l \leq (n-2)$
$$p + f(n-1, k-l+p) \leq f(n,k).$$
$k \geq 0$ can be bounded to anything as long as the induction step still works out.
Is there a function $f(n,k)$ that satisfies the above inequality such that $$f(n,0) \leq \frac{n^2-6n+6}{6}.$$
Notice that if we take $f(n,k) = n^2-k$, then this gives a solution, but I need the function $f(n,0)$ to be smaller.
| It seems the following.
Provided $n\ge 2$ put $k=l=\left\lfloor\frac{n-2}3\right\rfloor$ and
$p=\left\lfloor\frac{n-3}3\right\rfloor$. Then
$$\left\lfloor\frac{n-3}3\right\rfloor+f\left(n-1,\left\lfloor\frac{n-3}3\right\rfloor\right)\le f\left(n, \left\lfloor\frac{n-2}3\right\rfloor\right),$$
which should give a lower bound $f\left(n, \left\lfloor\frac{n-2}3\right\rfloor)\right)\ge\frac {n^2}{6}-O(n)$.
Now put $k=l=0$ and $p=\left\lfloor\frac{n-3}3\right\rfloor$.
Then $$f(n,0)\ge p+f(n-1,p)\ge\frac {n^2}{6}-O(n).$$
More precisely. The proposed lower bound is not far from your upper bound $f(n,0) \leq \frac{n^2-6n+6}{6}$. Indeed, for $n\ge 7$ we have
$$f(n,0)\ge \left\lfloor\frac{n-3}3\right\rfloor+f\left(n-1, \left\lfloor\frac{n-3}3\right\rfloor\right)\ge$$ $$\left\lfloor\frac{n-3}3\right\rfloor+\left\lfloor\frac{n-4}3\right\rfloor+f\left(n-2, \left\lfloor\frac{n-4}3\right\rfloor\right)\ge$$
$$\left\lfloor\frac{n-3}3\right\rfloor+\left\lfloor\frac{n-4}3\right\rfloor+\dots+\left\lfloor\frac{4}3\right\rfloor+f(2,0)\ge$$
$$-\frac{1}3+\sum_{k=4}^{n-3}\frac{k-1}3\ge \frac{n^2-7n+4}6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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optimize distance from point to plane with Lagrange method I am trying to optimize distance from point to plane using Lagrange multiplier.
Usually for such problems you are given specific point like (1,2,3) in 3D, and then an exact plane which is just the subject of Lagrange. But what I have here doesn't specify values for point and plane.
It says problem happens in a D-dimensional space. It denotes the point as X, the plane as (wT)x+b=0, where wT is the transpose of w (which is just a n-by-1 matrix I suppose), and requires me to use Lagrange to optimize the distance. The final result should be expressed with w, b and X. Without specific values I am really lost in how to approach such kind of problem. Any suggestions?
| I think what you are looking for is kind of this
Distance from point to plane
\begin{equation*}
D(\mathbf{x}_{0},\mathbf{P}) = \frac{\vert \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} \vert}
{\Vert \mathbf{w} \Vert_{2}}
\end{equation*}
Let
\begin{equation*}
\mathbf{P} = \lbrace \mathbf{x} \in \mathbb{R}^{n} / \mathbf{w}^{T} \mathbf{x} + \mathbf{b} = 0\rbrace
\end{equation*}
So
\begin{equation*}
D(\mathbf{x}_0, \mathbf{P}) = \inf_{\mathbf{x} \in \mathbf{P} } \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}
\end{equation*}
In order to get the minimum value satisfying above, it is needed to calculate de Lagragian
\begin{equation*}
\mathit{L}(\mathbf{x}, \lambda) = \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2} + \lambda ( \mathbf{w}^{T} \mathbf{x} + \mathbf{b} ) \hspace{0.3cm} \text{where} \hspace{0.3cm} \mathbf{x}^T \mathbf{x} = \Vert \mathbf{x} \Vert_{2}^{2}
\end{equation*}
Finding partial derivatives it yields:
\begin{equation*}
\frac{\partial \mathit{L}(\mathbf{x}, \lambda)}{\partial \mathbf{x}}
= \frac{\mathbf{x} - \mathbf{x}_{0}}{ \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}}
+ \lambda \mathbf{w} = 0
\end{equation*}
\begin{equation*}
\frac{\partial \mathit{L}(\mathbf{x}, \lambda)}{\partial \lambda}
= \mathbf{w}^{T} \mathbf{x} + \mathbf{b} = 0
\end{equation*}
Focusing on the second equation we have
\begin{equation*}
\mathbf{w}^{T} \mathbf{x} + \mathbf{b} = 0 \hspace{0.3cm} \text{so} \hspace{0.3cm}
\mathbf{w}^{T} ( \mathbf{x} - \mathbf{x}_{0} ) = - ( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )
\end{equation*}
In the first derivative we make a right internal product of $\mathbf{w}^{T}$
\begin{equation*}
\frac{\mathbf{w}^{T}(\mathbf{x} - \mathbf{x}_{0})}{ \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}}
+ \lambda \mathbf{w}^{T} \mathbf{w} = 0
\end{equation*}
from above, we have
\begin{equation*}
-\frac{( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )}{ \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}}
+ \lambda \mathbf{w}^{T} \mathbf{w} = 0
\end{equation*}
from which
\begin{equation*}
\lambda^{\star} = \frac{( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )}{ \Vert \mathbf{w} \Vert_{2}^{2} \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}}
\end{equation*}
Replacing $\lambda$ in the first equation, it yields;
\begin{equation*}
\frac{\mathbf{x} - \mathbf{x}_{0}}{ \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}}
+ \frac{( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )}{ \Vert \mathbf{w} \Vert_{2}^{2} \Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2}} \mathbf{w} = 0
\end{equation*}
So
\begin{equation*}
\mathbf{x} - \mathbf{x}_{0} = - \frac{( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )}{ \Vert \mathbf{w} \Vert_{2}^{2}} \mathbf{w} \hspace{0.3cm} \text{or}\hspace{0.3cm}
\mathbf{x}^{\star} = \mathbf{x}_{0} - \frac{( \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} )}{ \Vert \mathbf{w} \Vert_{2}^{2}} \mathbf{w}
\end{equation*}
Therefore
\begin{equation*}
\Vert \mathbf{x} - \mathbf{x}_{0} \Vert_{2} = \frac{\vert \mathbf{w}^{T} \mathbf{x}_{0} + \mathbf{b} \vert}{ \Vert \mathbf{w} \Vert_{2}}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a Polynomial in $x-\frac1x$ Given that $x^n - (1/x^n)$ is expressible as a polynomial in $x - (1/x)$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P(x-(1/x)) = x^5 - (1/x)^5.$
To start, I factored, giving:
$P(x-\frac{1}{x}) = (x-\frac{1}{x})(x^4+\frac{1}{x^4}+x^2+\frac{1}{x^2}+1).$
However, I cannot find a way to connect this to $x-\frac{1}{x}$. What should I do?
| $$\left(x^m-\dfrac1{x^m}\right)\left(x^2+\dfrac1{x^2}\right)=x^{m+2}-\dfrac1{x^{m+2}}+x^{m-2}-\dfrac1{x^{m-2}}$$
$$\implies F_{m+2}=F_m\left(x^2+\dfrac1{x^2}\right)+F_{m-2}$$ where $F_r=x^r-\dfrac1{x^r}$
Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=F_1^2+2$
$$F_3=\left(x-\dfrac1x\right)^3+3\left(x-\dfrac1x\right)$$
$$F_5=\left(\left(x-\dfrac1x\right)^2+2\right)F_3+F_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.